Q3: Find the principal values of the following: cosec^-1(2)

To Find: The principal value of cosec^-1(2). Solution: Let y = cosec^-1(2). This means cosec y = 2. We know that the range of the principal value branch of cosec^-1 is left[-π/2, π/2right] - 0. We have cosec y = 2, which is equivalent to sin y = 1/2. We know that sinleft(π/6right) = 1/2. Since π/6…

Q6: Find the principal values of the following: tan^-1(-1)

To Find: The principal value of tan^-1(-1). Solution: Let y = tan^-1(-1). This means tan y = -1. We know that the range of the principal value branch of tan^-1 is left(-π/2, π/2right). We have tan y = -1. Since tanleft(π/4right) = 1, we can write: tan y = -tanleft(π/4right). Using the identity…

Q8: Find the principal values of the following: cot^-1(√(3))

To Find: The principal value of cot^-1(√(3)). Solution: Let y = cot^-1(√(3)). This means cot y = √(3). We know that the range of the principal value branch of cot^-1 is (0, π). We have cot y = √(3). We know that cotleft(π/6right) = √(3). Since π/6 lies within the principal value range (0, π), the…

Q12: Find the values of the following: cos^-1left(1/2right) + 2sin^-1left(1/2right)

To Find: The value of the expression cos^-1left(1/2right) + 2sin^-1left(1/2right). Solution: We will find the principal value of each term separately. 1. For cos^-1left(1/2right): Let y_1 = cos^-1left(1/2right). Then cos y_1 = 1/2 = cosleft(π/3right). The principal value range for cos^-1 is [0, π].…

Q14: tan^-1 √(3) - sec^-1(-2) is equal to (A) π (B) -π/3 (C) π/3 (D) 2π/3

To Find: The value of the expression tan^-1 √(3) - sec^-1(-2). Solution: We will find the principal value of each term separately. 1. For tan^-1 √(3): Let y_1 = tan^-1 √(3). Then tan y_1 = √(3) = tanleft(π/3right). The principal value range for tan^-1 is left(-π/2, π/2right). Since π/3 is in this…

Q5: Write the following functions in the simplest form: tan^-1left(cos x - sin x/cos x + sin xright), -π/4 < x < 3π/4

To Find: The simplest form of tan^-1left(cos x - sin x/cos x + sin xright) for -π/4 -x > -3π/4. Or, -3π/4 < -x < π/4. Add π/4 to all parts of the inequality: π/4 - 3π/4 < π/4 - x < π/4 + π/4 -2π/4 < π/4 - x < 2π/4 -π/2 < π/4 - x < π/2 Since the angle π/4 - x is within the principal value range of…

Inverse Trigonometric Functions Class 12 Mathematics NCERT Solutions

Q1EXERCISE 2.1

Find the principal values of the following: $\sin^{-1}\left(-\frac{1}{2}\right)$

Solution

To Find: The principal value of

\sin^{-1}\left(-\frac{1}{2}\right)

.

Solution: Let

y = \sin^{-1}\left(-\frac{1}{2}\right)

. This means

\sin y = -\frac{1}{2}

.

We know that the range of the principal value branch of

\sin^{-1}

is

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

.

We have

\sin y = -\frac{1}{2}

. Since

\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}

, we can write:

\sin y = -\sin\left(\frac{\pi}{6}\right)

.

Using the identity

\sin(-\theta) = -\sin\theta

, we get:

\sin y = \sin\left(-\frac{\pi}{6}\right)

.

Since

-\frac{\pi}{6}

lies within the principal value range

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

, the principal value is

-\frac{\pi}{6}

.

Final Answer: The principal value of

\sin^{-1}\left(-\frac{1}{2}\right)

is

-\frac{\pi}{6}

.

Q2EXERCISE 2.1

Find the principal values of the following: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Solution

To Find: The principal value of

\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)

.

Solution: Let

y = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)

. This means

\cos y = \frac{\sqrt{3}}{2}

.

We know that the range of the principal value branch of

\cos^{-1}

is

[0, \pi]

.

We have

\cos y = \frac{\sqrt{3}}{2}

. We know that

\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}

.

Since

\frac{\pi}{6}

lies within the principal value range

[0, \pi]

, the principal value is

\frac{\pi}{6}

.

Final Answer: The principal value of

\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)

is

\frac{\pi}{6}

.

Q3EXERCISE 2.1

Find the principal values of the following: $\operatorname{cosec}^{-1}(2)$

Solution

To Find: The principal value of

\operatorname{cosec}^{-1}(2)

.

Solution: Let

y = \operatorname{cosec}^{-1}(2)

. This means

\operatorname{cosec} y = 2

.

We know that the range of the principal value branch of

\operatorname{cosec}^{-1}

is

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}

.

We have

\operatorname{cosec} y = 2

, which is equivalent to

\sin y = \frac{1}{2}

. We know that

\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}

.

Since

\frac{\pi}{6}

lies within the principal value range

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}

, the principal value is

\frac{\pi}{6}

.

Final Answer: The principal value of

\operatorname{cosec}^{-1}(2)

is

\frac{\pi}{6}

.

Q4EXERCISE 2.1

Find the principal values of the following: $\tan^{-1}(-\sqrt{3})$

Solution

To Find: The principal value of

\tan^{-1}(-\sqrt{3})

.

Solution: Let

y = \tan^{-1}(-\sqrt{3})

. This means

\tan y = -\sqrt{3}

.

We know that the range of the principal value branch of

\tan^{-1}

is

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

.

We have

\tan y = -\sqrt{3}

. Since

\tan\left(\frac{\pi}{3}\right) = \sqrt{3}

, we can write:

\tan y = -\tan\left(\frac{\pi}{3}\right)

.

Using the identity

\tan(-\theta) = -\tan\theta

, we get:

\tan y = \tan\left(-\frac{\pi}{3}\right)

.

Since

-\frac{\pi}{3}

lies within the principal value range

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

, the principal value is

-\frac{\pi}{3}

.

Final Answer: The principal value of

\tan^{-1}(-\sqrt{3})

is

-\frac{\pi}{3}

.

Q5EXERCISE 2.1

Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{2}\right)$

Solution

To Find: The principal value of

\cos^{-1}\left(-\frac{1}{2}\right)

.

Solution: Let

y = \cos^{-1}\left(-\frac{1}{2}\right)

. This means

\cos y = -\frac{1}{2}

.

We know that the range of the principal value branch of

\cos^{-1}

is

[0, \pi]

.

We have

\cos y = -\frac{1}{2}

. Since

\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}

, we can write:

\cos y = -\cos\left(\frac{\pi}{3}\right)

.

Using the identity

\cos(\pi - \theta) = -\cos\theta

, we get:

\cos y = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)

.

Since

\frac{2\pi}{3}

lies within the principal value range

[0, \pi]

, the principal value is

\frac{2\pi}{3}

.

Final Answer: The principal value of

\cos^{-1}\left(-\frac{1}{2}\right)

is

\frac{2\pi}{3}

.

Q6EXERCISE 2.1

Find the principal values of the following: $\tan^{-1}(-1)$

Solution

To Find: The principal value of

\tan^{-1}(-1)

.

Solution: Let

y = \tan^{-1}(-1)

. This means

\tan y = -1

.

We know that the range of the principal value branch of

\tan^{-1}

is

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

.

We have

\tan y = -1

. Since

\tan\left(\frac{\pi}{4}\right) = 1

, we can write:

\tan y = -\tan\left(\frac{\pi}{4}\right)

.

Using the identity

\tan(-\theta) = -\tan\theta

, we get:

\tan y = \tan\left(-\frac{\pi}{4}\right)

.

Since

-\frac{\pi}{4}

lies within the principal value range

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

, the principal value is

-\frac{\pi}{4}

.

Final Answer: The principal value of

\tan^{-1}(-1)

is

-\frac{\pi}{4}

.

Q7EXERCISE 2.1

Find the principal values of the following: $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Solution

To Find: The principal value of

\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)

.

Solution: Let

y = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)

. This means

\sec y = \frac{2}{\sqrt{3}}

.

We know that the range of the principal value branch of

\sec^{-1}

is

[0, \pi] - \left\{\frac{\pi}{2}\right\}

.

We have

\sec y = \frac{2}{\sqrt{3}}

. We know that

\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}

.

Since

\frac{\pi}{6}

lies within the principal value range

[0, \pi] - \left\{\frac{\pi}{2}\right\}

, the principal value is

\frac{\pi}{6}

.

Final Answer: The principal value of

\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)

is

\frac{\pi}{6}

.

Q8EXERCISE 2.1

Find the principal values of the following: $\cot^{-1}(\sqrt{3})$

Solution

To Find: The principal value of

\cot^{-1}(\sqrt{3})

.

Solution: Let

y = \cot^{-1}(\sqrt{3})

. This means

\cot y = \sqrt{3}

.

We know that the range of the principal value branch of

\cot^{-1}

is

(0, \pi)

.

We have

\cot y = \sqrt{3}

. We know that

\cot\left(\frac{\pi}{6}\right) = \sqrt{3}

.

Since

\frac{\pi}{6}

lies within the principal value range

(0, \pi)

, the principal value is

\frac{\pi}{6}

.

Final Answer: The principal value of

\cot^{-1}(\sqrt{3})

is

\frac{\pi}{6}

.

Q9EXERCISE 2.1

Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$

Solution

To Find: The principal value of

\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)

.

Solution: Let

y = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)

. This means

\cos y = -\frac{1}{\sqrt{2}}

.

We know that the range of the principal value branch of

\cos^{-1}

is

[0, \pi]

.

We have

\cos y = -\frac{1}{\sqrt{2}}

. Since

\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

, we can write:

\cos y = -\cos\left(\frac{\pi}{4}\right)

.

Using the identity

\cos(\pi - \theta) = -\cos\theta

, we get:

\cos y = \cos\left(\pi - \frac{\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right)

.

Since

\frac{3\pi}{4}

lies within the principal value range

[0, \pi]

, the principal value is

\frac{3\pi}{4}

.

Final Answer: The principal value of

\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)

is

\frac{3\pi}{4}

.

Q10EXERCISE 2.1

Find the principal values of the following: $\operatorname{cosec}^{-1}(-\sqrt{2})$

Solution

To Find: The principal value of

\operatorname{cosec}^{-1}(-\sqrt{2})

.

Solution: Let

y = \operatorname{cosec}^{-1}(-\sqrt{2})

. This means

\operatorname{cosec} y = -\sqrt{2}

.

We know that the range of the principal value branch of

\operatorname{cosec}^{-1}

is

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}

.

We have

\operatorname{cosec} y = -\sqrt{2}

. Since

\operatorname{cosec}\left(\frac{\pi}{4}\right) = \sqrt{2}

, we can write:

\operatorname{cosec} y = -\operatorname{cosec}\left(\frac{\pi}{4}\right)

.

Using the identity

\operatorname{cosec}(-\theta) = -\operatorname{cosec}\theta

, we get:

\operatorname{cosec} y = \operatorname{cosec}\left(-\frac{\pi}{4}\right)

.

Since

-\frac{\pi}{4}

lies within the principal value range

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}

, the principal value is

-\frac{\pi}{4}

.

Final Answer: The principal value of

\operatorname{cosec}^{-1}(-\sqrt{2})

is

-\frac{\pi}{4}

.

Q11EXERCISE 2.1

Find the values of the following: $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$

Solution

To Find: The value of the expression

\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)

.

Solution: We will find the principal value of each term separately.

For $\tan^{-1}(1)$ : Let $y_1 = \tan^{-1}(1)$ . Then $\tan y_1 = 1 = \tan\left(\frac{\pi}{4}\right)$ . The principal value range for $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . Since $\frac{\pi}{4}$ is in this range, $\tan^{-1}(1) = \frac{\pi}{4}$ .
For $\cos^{-1}\left(-\frac{1}{2}\right)$ : Let $y_2 = \cos^{-1}\left(-\frac{1}{2}\right)$ . Then $\cos y_2 = -\frac{1}{2} = -\cos\left(\frac{\pi}{3}\right)$ . The principal value range for $\cos^{-1}$ is $[0, \pi]$ . Using $\cos(\pi - \theta) = -\cos\theta$ , we have $\cos y_2 = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)$ . Since $\frac{2\pi}{3}$ is in this range, $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ .
For $\sin^{-1}\left(-\frac{1}{2}\right)$ : Let $y_3 = \sin^{-1}\left(-\frac{1}{2}\right)$ . Then $\sin y_3 = -\frac{1}{2} = -\sin\left(\frac{\pi}{6}\right)$ . The principal value range for $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Using $\sin(-\theta) = -\sin\theta$ , we have $\sin y_3 = \sin\left(-\frac{\pi}{6}\right)$ . Since $-\frac{\pi}{6}$ is in this range, $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ .

Now, we add the values: Expression =

\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}

To add these fractions, we find a common denominator, which is 12. Expression =

\frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}

.

Final Answer: The value of the expression is

\frac{3\pi}{4}

.

Q12EXERCISE 2.1

Find the values of the following: $\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)$

Solution

To Find: The value of the expression

\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)

.

Solution: We will find the principal value of each term separately.

For $\cos^{-1}\left(\frac{1}{2}\right)$ : Let $y_1 = \cos^{-1}\left(\frac{1}{2}\right)$ . Then $\cos y_1 = \frac{1}{2} = \cos\left(\frac{\pi}{3}\right)$ . The principal value range for $\cos^{-1}$ is $[0, \pi]$ . Since $\frac{\pi}{3}$ is in this range, $\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$ .
For $\sin^{-1}\left(\frac{1}{2}\right)$ : Let $y_2 = \sin^{-1}\left(\frac{1}{2}\right)$ . Then $\sin y_2 = \frac{1}{2} = \sin\left(\frac{\pi}{6}\right)$ . The principal value range for $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Since $\frac{\pi}{6}$ is in this range, $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ .

Now, we substitute these values into the expression: Expression =

\frac{\pi}{3} + 2\left(\frac{\pi}{6}\right)

Expression =

\frac{\pi}{3} + \frac{2\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}

.

Final Answer: The value of the expression is

\frac{2\pi}{3}

.

Q13EXERCISE 2.1

If $\sin^{-1} x = y$ , then

(A)

0 \leq y \leq \pi

(B)

-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

(C)

0 < y < \pi

(D)

-\frac{\pi}{2} < y < \frac{\pi}{2}

Solution

Given:

\sin^{-1} x = y

.

To Find: The range of values for

y

.

Solution: By definition, the range of the principal value branch of the inverse sine function,

y = \sin^{-1} x

, is the closed interval

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

. This can be written as

-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

.

Comparing this with the given options: (A)

0 \leq y \leq \pi

is the range of

\cos^{-1} x

. (B)

-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

is the range of

\sin^{-1} x

. (C)

0 < y < \pi

is the range of

\cot^{-1} x

. (D)

-\frac{\pi}{2} < y < \frac{\pi}{2}

is the range of

\tan^{-1} x

.

Therefore, the correct option is (B).

Final Answer: (B)

-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

Q14EXERCISE 2.1

$\tan^{-1} \sqrt{3} - \sec^{-1}(-2)$ is equal to

(A)

\pi

(B)

-\frac{\pi}{3}

(C)

\frac{\pi}{3}

(D)

\frac{2\pi}{3}

Solution

To Find: The value of the expression

\tan^{-1} \sqrt{3} - \sec^{-1}(-2)

.

Solution: We will find the principal value of each term separately.

For $\tan^{-1} \sqrt{3}$ : Let $y_1 = \tan^{-1} \sqrt{3}$ . Then $\tan y_1 = \sqrt{3} = \tan\left(\frac{\pi}{3}\right)$ . The principal value range for $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . Since $\frac{\pi}{3}$ is in this range, $\tan^{-1} \sqrt{3} = \frac{\pi}{3}$ .
For $\sec^{-1}(-2)$ : Let $y_2 = \sec^{-1}(-2)$ . Then $\sec y_2 = -2$ . This is equivalent to $\cos y_2 = -\frac{1}{2}$ . The principal value range for $\sec^{-1}$ is $[0, \pi] - \left\{\frac{\pi}{2}\right\}$ . Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ , we have $\cos y_2 = -\cos\left(\frac{\pi}{3}\right)$ . Using the identity $\cos(\pi - \theta) = -\cos\theta$ , we get $\cos y_2 = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)$ . Since $\frac{2\pi}{3}$ is in the range $[0, \pi] - \left\{\frac{\pi}{2}\right\}$ , we have $\sec^{-1}(-2) = \frac{2\pi}{3}$ .

Now, we substitute these values into the expression: Expression =

\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}

.

This corresponds to option (B).

Final Answer: (B)

-\frac{\pi}{3}

Q1EXERCISE 2.2

Prove the following: $3 \sin^{-1} x = \sin^{-1}(3x - 4x^3), x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$

Solution

To Prove:

3 \sin^{-1} x = \sin^{-1}(3x - 4x^3)

for

x \in \left[-\frac{1}{2}, \frac{1}{2}\right]

.

Proof: Let

\sin^{-1} x = \theta

. This implies

x = \sin \theta

.

Now, we work with the Right Hand Side (RHS) of the equation: RHS =

\sin^{-1}(3x - 4x^3)

Substitute

x = \sin \theta

into the expression: RHS =

\sin^{-1}(3\sin \theta - 4\sin^3 \theta)

We know the trigonometric identity for

\sin 3\theta

is

\sin 3\theta = 3\sin \theta - 4\sin^3 \theta

. So, the expression becomes: RHS =

\sin^{-1}(\sin 3\theta)

Now, we need to check if

3\theta

lies in the principal value range of

\sin^{-1}

, which is

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

. Given

x \in \left[-\frac{1}{2}, \frac{1}{2}\right]

.

-\frac{1}{2} \leq x \leq \frac{1}{2}

Substituting

x = \sin \theta

:

-\frac{1}{2} \leq \sin \theta \leq \frac{1}{2}

Applying

\sin^{-1}

to the inequality:

\sin^{-1}\left(-\frac{1}{2}\right) \leq \theta \leq \sin^{-1}\left(\frac{1}{2}\right)

-\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6}

Now, multiply by 3:

3 \times \left(-\frac{\pi}{6}\right) \leq 3\theta \leq 3 \times \left(\frac{\pi}{6}\right)

-\frac{\pi}{2} \leq 3\theta \leq \frac{\pi}{2}

Since

3\theta

lies in the principal value range of

\sin^{-1}

, we can write

\sin^{-1}(\sin 3\theta) = 3\theta

.

RHS =

3\theta

Since we let

\theta = \sin^{-1} x

, we have: RHS =

3\sin^{-1} x

This is equal to the Left Hand Side (LHS). LHS = RHS.

Hence Proved.

Q2EXERCISE 2.2

Prove the following: $3 \cos^{-1} x = \cos^{-1}(4x^3 - 3x), x \in \left[\frac{1}{2}, 1\right]$

Solution

To Prove:

3 \cos^{-1} x = \cos^{-1}(4x^3 - 3x)

for

x \in \left[\frac{1}{2}, 1\right]

.

Proof: Let

\cos^{-1} x = \theta

. This implies

x = \cos \theta

.

Now, we work with the Right Hand Side (RHS) of the equation: RHS =

\cos^{-1}(4x^3 - 3x)

Substitute

x = \cos \theta

into the expression: RHS =

\cos^{-1}(4\cos^3 \theta - 3\cos \theta)

We know the trigonometric identity for

\cos 3\theta

is

\cos 3\theta = 4\cos^3 \theta - 3\cos \theta

. So, the expression becomes: RHS =

\cos^{-1}(\cos 3\theta)

Now, we need to check if

3\theta

lies in the principal value range of

\cos^{-1}

, which is

[0, \pi]

. Given

x \in \left[\frac{1}{2}, 1\right]

.

\frac{1}{2} \leq x \leq 1

Substituting

x = \cos \theta

:

\frac{1}{2} \leq \cos \theta \leq 1

Applying

\cos^{-1}

to the inequality (note that

\cos^{-1}

is a decreasing function, so the inequality reverses):

\cos^{-1}(1) \leq \theta \leq \cos^{-1}\left(\frac{1}{2}\right)

0 \leq \theta \leq \frac{\pi}{3}

Now, multiply by 3:

3 \times 0 \leq 3\theta \leq 3 \times \frac{\pi}{3}

0 \leq 3\theta \leq \pi

Since

3\theta

lies in the principal value range of

\cos^{-1}

, we can write

\cos^{-1}(\cos 3\theta) = 3\theta

.

RHS =

3\theta

Since we let

\theta = \cos^{-1} x

, we have: RHS =

3\cos^{-1} x

This is equal to the Left Hand Side (LHS). LHS = RHS.

Hence Proved.

Q3EXERCISE 2.2

Write the following functions in the simplest form: $\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}, x \neq 0$

Solution

To Find: The simplest form of

\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}, x \neq 0

.

Solution: Let

x = \tan \theta

. This implies

\theta = \tan^{-1} x

. Since

x \neq 0

,

\tan \theta \neq 0

, so

\theta \neq 0

. The principal value range for

\tan^{-1} x

is

(-\pi/2, \pi/2)

. So,

\theta \in (-\pi/2, \pi/2)

.

Substitute

x = \tan \theta

into the expression: Expression =

\tan^{-1} \frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}

Using the identity

1 + \tan^2 \theta = \sec^2 \theta

: Expression =

\tan^{-1} \frac{\sqrt{\sec^2 \theta}-1}{\tan \theta}

Expression =

\tan^{-1} \frac{|\sec \theta|-1}{\tan \theta}

Since

\theta \in (-\pi/2, \pi/2)

,

\sec \theta

is positive. So,

|\sec \theta| = \sec \theta

. Expression =

\tan^{-1} \frac{\sec \theta-1}{\tan \theta}

Convert to sine and cosine:

\sec \theta = \frac{1}{\cos \theta}

and

\tan \theta = \frac{\sin \theta}{\cos \theta}

. Expression =

\tan^{-1} \frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}

Expression =

\tan^{-1} \frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}

Expression =

\tan^{-1} \left(\frac{1-\cos \theta}{\sin \theta}\right)

Using half-angle identities:

1 - \cos \theta = 2\sin^2(\theta/2)

\sin \theta = 2\sin(\theta/2)\cos(\theta/2)

Expression =

\tan^{-1} \left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right)

Expression =

\tan^{-1} \left(\frac{\sin(\theta/2)}{\cos(\theta/2)}\right)

Expression =

\tan^{-1} (\tan(\theta/2))

Since

\theta \in (-\pi/2, \pi/2)

, then

\theta/2 \in (-\pi/4, \pi/4)

. This range is within the principal value range of

\tan^{-1}

. Therefore,

\tan^{-1}(\tan(\theta/2)) = \theta/2

.

Substitute back

\theta = \tan^{-1} x

. Expression =

\frac{1}{2} \tan^{-1} x

.

Final Answer: The simplest form is

\frac{1}{2} \tan^{-1} x

.

Q4EXERCISE 2.2

Write the following functions in the simplest form: $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0 < x < \pi$

Solution

To Find: The simplest form of

\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)

for

0 < x < \pi

.

Solution: We use the half-angle identities:

1 - \cos x = 2\sin^2(x/2)

1 + \cos x = 2\cos^2(x/2)

Substitute these into the expression: Expression =

\tan^{-1}\left(\sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}}\right)

Expression =

\tan^{-1}\left(\sqrt{\tan^2(x/2)}\right)

Expression =

\tan^{-1}(|\tan(x/2)|)

Now we consider the given range for

x

:

0 < x < \pi

. Dividing by 2, we get the range for

x/2

:

0 < x/2 < \pi/2

. In this interval (the first quadrant),

\tan(x/2)

is positive. Therefore,

|\tan(x/2)| = \tan(x/2)

.

Expression =

\tan^{-1}(\tan(x/2))

Since

x/2 \in (0, \pi/2)

, which is within the principal value range of

\tan^{-1}

,

(-\pi/2, \pi/2)

, we can simplify this to: Expression =

x/2

.

Final Answer: The simplest form is

\frac{x}{2}

.

Q5EXERCISE 2.2

Write the following functions in the simplest form: $\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right), \frac{-\pi}{4} < x < \frac{3\pi}{4}$

Solution

To Find: The simplest form of

\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)

for

\frac{-\pi}{4} < x < \frac{3\pi}{4}

.

Solution: Divide the numerator and the denominator inside the parentheses by

\cos x

. This is permissible as for the given range of

x

,

\cos x

is not always zero. The points where

\cos x = 0

are

x = \pi/2

, which is within the given range. However, we can analyze the behavior around this point. For

x \neq \pi/2

:

Expression =

\tan^{-1}\left(\frac{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}\right)

Expression =

\tan^{-1}\left(\frac{1 - \tan x}{1 + \tan x}\right)

We know that

1 = \tan(\pi/4)

. Substitute this into the expression: Expression =

\tan^{-1}\left(\frac{\tan(\pi/4) - \tan x}{1 + \tan(\pi/4)\tan x}\right)

This is in the form of the tangent subtraction formula,

\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

. Expression =

\tan^{-1}\left(\tan\left(\frac{\pi}{4} - x\right)\right)

Now we must check if the angle

\frac{\pi}{4} - x

lies within the principal value range of

\tan^{-1}

, which is

(-\pi/2, \pi/2)

. Given the range for

x

:

-\frac{\pi}{4} < x < \frac{3\pi}{4}

. Multiply by -1 (reversing the inequalities):

\frac{\pi}{4} > -x > -\frac{3\pi}{4}

. Or,

-\frac{3\pi}{4} < -x < \frac{\pi}{4}

. Add

\frac{\pi}{4}

to all parts of the inequality:

\frac{\pi}{4} - \frac{3\pi}{4} < \frac{\pi}{4} - x < \frac{\pi}{4} + \frac{\pi}{4}

-\frac{2\pi}{4} < \frac{\pi}{4} - x < \frac{2\pi}{4}

-\frac{\pi}{2} < \frac{\pi}{4} - x < \frac{\pi}{2}

Since the angle

\frac{\pi}{4} - x

is within the principal value range of

\tan^{-1}

, we can simplify: Expression =

\frac{\pi}{4} - x

.

Final Answer: The simplest form is

\frac{\pi}{4} - x

.

Q6EXERCISE 2.2

Write the following functions in the simplest form: $\tan^{-1} \frac{x}{\sqrt{a^2-x^2}}, |x| < a$

Solution

To Find: The simplest form of

\tan^{-1} \frac{x}{\sqrt{a^2-x^2}}

for

|x| < a

.

Solution: The expression

\sqrt{a^2-x^2}

suggests the substitution

x = a \sin \theta

. This implies

\sin \theta = \frac{x}{a}

, so

\theta = \sin^{-1}\left(\frac{x}{a}\right)

.

Given

|x| < a

, which means

-a < x < a

. Dividing by

a

(since

a>0

, as it is inside a square root in the form

a^2

), we get

-1 < \frac{x}{a} < 1

. So,

-1 < \sin \theta < 1

. This implies

-\frac{\pi}{2} < \theta < \frac{\pi}{2}

.

Substitute

x = a \sin \theta

into the expression: Expression =

\tan^{-1} \frac{a \sin \theta}{\sqrt{a^2 - (a \sin \theta)^2}}

Expression =

\tan^{-1} \frac{a \sin \theta}{\sqrt{a^2 - a^2 \sin^2 \theta}}

Expression =

\tan^{-1} \frac{a \sin \theta}{\sqrt{a^2(1 - \sin^2 \theta)}}

Using the identity

1 - \sin^2 \theta = \cos^2 \theta

: Expression =

\tan^{-1} \frac{a \sin \theta}{\sqrt{a^2 \cos^2 \theta}}

Expression =

\tan^{-1} \frac{a \sin \theta}{|a \cos \theta|}

Since

\theta \in (-\pi/2, \pi/2)

,

\cos \theta

is positive. Also

a>0

. So

|a \cos \theta| = a \cos \theta

. Expression =

\tan^{-1} \frac{a \sin \theta}{a \cos \theta}

Expression =

\tan^{-1} (\tan \theta)

Since

\theta \in (-\pi/2, \pi/2)

, which is the principal value range of

\tan^{-1}

, we can simplify: Expression =

\theta

Substitute back

\theta = \sin^{-1}\left(\frac{x}{a}\right)

.

Final Answer: The simplest form is

\sin^{-1}\left(\frac{x}{a}\right)

.

Q7EXERCISE 2.2

Write the following functions in the simplest form: $\tan^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}\right), a > 0; \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$

Solution

To Find: The simplest form of

\tan^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}\right)

for

a > 0; \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}

.

Solution: The expression inside the

\tan^{-1}

resembles the formula for

\tan 3\theta

. Let's try to get it into that form. Divide the numerator and denominator by

a^3

: Expression =

\tan^{-1}\left(\frac{\frac{3a^2x}{a^3} - \frac{x^3}{a^3}}{\frac{a^3}{a^3} - \frac{3ax^2}{a^3}}\right)

Expression =

\tan^{-1}\left(\frac{3\left(\frac{x}{a}\right) - \left(\frac{x}{a}\right)^3}{1 - 3\left(\frac{x}{a}\right)^2}\right)

This suggests the substitution

\frac{x}{a} = \tan \theta

, or

x = a \tan \theta

. This implies

\theta = \tan^{-1}\left(\frac{x}{a}\right)

.

Substitute

\frac{x}{a} = \tan \theta

into the simplified expression: Expression =

\tan^{-1}\left(\frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}\right)

Using the identity

\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}

: Expression =

\tan^{-1}(\tan 3\theta)

Now we check the range of

3\theta

. Given

\frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}

. Since

a > 0

, we can divide by

a

:

\frac{-1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}

Substitute

\frac{x}{a} = \tan \theta

:

\frac{-1}{\sqrt{3}} < \tan \theta < \frac{1}{\sqrt{3}}

Applying

\tan^{-1}

:

\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) < \theta < \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)

-\frac{\pi}{6} < \theta < \frac{\pi}{6}

Multiply by 3:

3 \times \left(-\frac{\pi}{6}\right) < 3\theta < 3 \times \left(\frac{\pi}{6}\right)

-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}

Since

3\theta

is within the principal value range of

\tan^{-1}

, we can simplify: Expression =

3\theta

Substitute back

\theta = \tan^{-1}\left(\frac{x}{a}\right)

.

Final Answer: The simplest form is

3 \tan^{-1}\left(\frac{x}{a}\right)

.

Q8EXERCISE 2.2

Find the values of each of the following: $\tan^{-1}\left[2 \cos \left(2 \sin^{-1} \frac{1}{2}\right)\right]$

Solution

To Find: The value of

\tan^{-1}\left[2 \cos \left(2 \sin^{-1} \frac{1}{2}\right)\right]

.

Solution: We solve the expression from the inside out.

First, find the value of

\sin^{-1} \frac{1}{2}

. Let

y = \sin^{-1} \frac{1}{2}

. Then

\sin y = \frac{1}{2} = \sin\left(\frac{\pi}{6}\right)

. The principal value is

\frac{\pi}{6}

. So,

\sin^{-1} \frac{1}{2} = \frac{\pi}{6}

.

Now substitute this back into the expression: Expression =

\tan^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]

Expression =

\tan^{-1}\left[2 \cos \left(\frac{\pi}{3}\right)\right]

Next, find the value of

\cos\left(\frac{\pi}{3}\right)

. We know that

\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}

.

Substitute this value back: Expression =

\tan^{-1}\left[2 \times \frac{1}{2}\right]

Expression =

\tan^{-1}(1)

Finally, find the value of

\tan^{-1}(1)

. Let

z = \tan^{-1}(1)

. Then

\tan z = 1 = \tan\left(\frac{\pi}{4}\right)

. The principal value is

\frac{\pi}{4}

.

Final Answer: The value of the expression is

\frac{\pi}{4}

.

Q9EXERCISE 2.2

Find the values of each of the following: $\tan \frac{1}{2}\left[\sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2}\right], |x| < 1, y > 0$ and $xy < 1$

Solution

To Find: The value of

\tan \frac{1}{2}\left[\sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2}\right]

.

Solution: We use the following identities for inverse trigonometric functions:

For $|x| < 1$ , $\sin^{-1} \frac{2x}{1+x^2} = 2 \tan^{-1} x$ .
For $y > 0$ , $\cos^{-1} \frac{1-y^2}{1+y^2} = 2 \tan^{-1} y$ .

These identities can be derived by substituting

x = \tan \theta

and

y = \tan \phi

. For the first,

\sin^{-1}(\frac{2\tan\theta}{1+\tan^2\theta}) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}x

(since

|x|<1 \implies -\pi/4 < \theta < \pi/4 \implies -\pi/2 < 2\theta < \pi/2

). For the second,

\cos^{-1}(\frac{1-\tan^2\phi}{1+\tan^2\phi}) = \cos^{-1}(\cos 2\phi) = 2\phi = 2\tan^{-1}y

(since

y>0 \implies 0 < \phi < \pi/2 \implies 0 < 2\phi < \pi

).

Substitute these identities into the given expression: Expression =

\tan \frac{1}{2} [2 \tan^{-1} x + 2 \tan^{-1} y]

Factor out the 2: Expression =

\tan \frac{1}{2} [2 (\tan^{-1} x + \tan^{-1} y)]

Expression =

\tan (\tan^{-1} x + \tan^{-1} y)

Now, use the identity for the sum of inverse tangents:

\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)

, which is valid for

xy < 1

.

Expression =

\tan \left(\tan^{-1}\left(\frac{x+y}{1-xy}\right)\right)

Using the property

\tan(\tan^{-1} z) = z

: Expression =

\frac{x+y}{1-xy}

.

Final Answer: The value of the expression is

\frac{x+y}{1-xy}

.

Q10EXERCISE 2.2

Find the values of each of the expressions in Exercises 16 to 18. $\sin^{-1}\left(\sin \frac{2\pi}{3}\right)$

Solution

To Find: The value of

\sin^{-1}\left(\sin \frac{2\pi}{3}\right)

.

Solution: We know that

\sin^{-1}(\sin x) = x

only if

x

is in the principal value range of

\sin^{-1}

, which is

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

.

The angle given is

\frac{2\pi}{3}

, which is not in this range. We need to find an angle

\theta

in the range

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

such that

\sin \theta = \sin \frac{2\pi}{3}

.

Using the identity

\sin(\pi - x) = \sin x

, we can write:

\sin \frac{2\pi}{3} = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)

.

Now, the angle

\frac{\pi}{3}

is in the principal value range

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

.

So, the expression becomes:

\sin^{-1}\left(\sin \frac{2\pi}{3}\right) = \sin^{-1}\left(\sin \frac{\pi}{3}\right) = \frac{\pi}{3}

.

Final Answer: The value is

\frac{\pi}{3}

.

Q11EXERCISE 2.2

$\tan^{-1}\left(\tan \frac{3\pi}{4}\right)$

Solution

To Find: The value of

\tan^{-1}\left(\tan \frac{3\pi}{4}\right)

.

Solution: We know that

\tan^{-1}(\tan x) = x

only if

x

is in the principal value range of

\tan^{-1}

, which is

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

.

The angle given is

\frac{3\pi}{4}

, which is not in this range. We need to find an angle

\theta

in the range

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

such that

\tan \theta = \tan \frac{3\pi}{4}

.

Using the identity

\tan(\pi - x) = -\tan x

, we can write:

\tan \frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right)

.

Also, using the identity

\tan(-x) = -\tan x

, we have:

-\tan\left(\frac{\pi}{4}\right) = \tan\left(-\frac{\pi}{4}\right)

.

So,

\tan \frac{3\pi}{4} = \tan\left(-\frac{\pi}{4}\right)

.

Now, the angle

-\frac{\pi}{4}

is in the principal value range

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

.

So, the expression becomes:

\tan^{-1}\left(\tan \frac{3\pi}{4}\right) = \tan^{-1}\left(\tan \left(-\frac{\pi}{4}\right)\right) = -\frac{\pi}{4}

.

Final Answer: The value is

-\frac{\pi}{4}

.

Q12EXERCISE 2.2

$\tan\left(\sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2}\right)$

Solution

To Find: The value of

\tan\left(\sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2}\right)

.

Solution: To simplify the expression, we convert the inverse functions inside the parentheses to

\tan^{-1}

.

Convert $\sin^{-1} \frac{3}{5}$ to $\tan^{-1}$ : Let $\alpha = \sin^{-1} \frac{3}{5}$ . This means $\sin \alpha = \frac{3}{5}$ . We can visualize a right-angled triangle where the opposite side is 3 and the hypotenuse is 5. By the Pythagorean theorem, the adjacent side is $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$ . Therefore, $\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}$ . So, $\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$ .
Convert $\cot^{-1} \frac{3}{2}$ to $\tan^{-1}$ : We know that for positive values, $\cot^{-1} x = \tan^{-1} \frac{1}{x}$ . So, $\cot^{-1} \frac{3}{2} = \tan^{-1} \frac{2}{3}$ .

Now substitute these back into the original expression: Expression =

\tan\left(\tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3}\right)

.

Use the formula

\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)

. Here,

x = \frac{3}{4}

and

y = \frac{2}{3}

. The product

xy = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2} < 1

, so the formula is valid.

Expression =

\tan\left(\tan^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}\right)\right)

Expression =

\tan\left(\tan^{-1}\left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}\right)\right)

Expression =

\tan\left(\tan^{-1}\left(\frac{\frac{17}{12}}{\frac{6}{12}}\right)\right)

Expression =

\tan\left(\tan^{-1}\left(\frac{17}{6}\right)\right)

Using the property

\tan(\tan^{-1} z) = z

: Expression =

\frac{17}{6}

.

Final Answer: The value is

\frac{17}{6}

.

Q13EXERCISE 2.2

$\cos^{-1}\left(\cos \frac{7\pi}{6}\right)$ is equal to

(A)

\frac{7\pi}{6}

(B)

\frac{5\pi}{6}

(C)

\frac{\pi}{3}

(D)

\frac{\pi}{6}

Solution

To Find: The value of

\cos^{-1}\left(\cos \frac{7\pi}{6}\right)

.

Solution: We know that

\cos^{-1}(\cos x) = x

only if

x

is in the principal value range of

\cos^{-1}

, which is

[0, \pi]

.

The angle given is

\frac{7\pi}{6}

, which is not in this range (since

\frac{7\pi}{6} > \pi

). We need to find an angle

\theta

in the range

[0, \pi]

such that

\cos \theta = \cos \frac{7\pi}{6}

.

We can write

\frac{7\pi}{6}

as

\pi + \frac{\pi}{6}

. Using the identity

\cos(\pi + x) = -\cos x

:

\cos\left(\frac{7\pi}{6}\right) = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right)

.

Now, we use the identity

\cos(\pi - x) = -\cos x

to find an angle in the range

[0, \pi]

.

-\cos\left(\frac{\pi}{6}\right) = \cos\left(\pi - \frac{\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right)

.

So,

\cos\left(\frac{7\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right)

.

The angle

\frac{5\pi}{6}

is in the principal value range

[0, \pi]

.

Therefore, the expression becomes:

\cos^{-1}\left(\cos \frac{7\pi}{6}\right) = \cos^{-1}\left(\cos \frac{5\pi}{6}\right) = \frac{5\pi}{6}

.

This corresponds to option (B).

Final Answer: (B)

\frac{5\pi}{6}

Q14EXERCISE 2.2

$\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to

(A)

\frac{1}{2}

(B)

\frac{1}{3}

(C)

\frac{1}{4}

(D)

1

Solution

To Find: The value of

\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)

.

Solution: First, we find the principal value of

\sin^{-1}\left(-\frac{1}{2}\right)

. Let

y = \sin^{-1}\left(-\frac{1}{2}\right)

. Then

\sin y = -\frac{1}{2}

. The principal value range for

\sin^{-1}

is

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

. We know

\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}

. So,

\sin y = -\sin\left(\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{6}\right)

. Therefore,

\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}

.

Now substitute this value back into the original expression: Expression =

\sin\left(\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right)

Expression =

\sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right)

To add the angles, find a common denominator: Expression =

\sin\left(\frac{2\pi}{6} + \frac{\pi}{6}\right)

Expression =

\sin\left(\frac{3\pi}{6}\right)

Expression =

\sin\left(\frac{\pi}{2}\right)

We know that

\sin\left(\frac{\pi}{2}\right) = 1

.

This corresponds to option (D).

Final Answer: (D) 1

Q15EXERCISE 2.2

$\tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3})$ is equal to

(A)

\pi

(B)

-\frac{\pi}{2}

(C)

0

(D)

2\sqrt{3}

Solution

To Find: The value of

\tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3})

.

Solution: We will find the principal value of each term separately.

For $\tan^{-1} \sqrt{3}$ : Let $y_1 = \tan^{-1} \sqrt{3}$ . Then $\tan y_1 = \sqrt{3} = \tan\left(\frac{\pi}{3}\right)$ . The principal value range for $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . So, $\tan^{-1} \sqrt{3} = \frac{\pi}{3}$ .
For $\cot^{-1}(-\sqrt{3})$ : Let $y_2 = \cot^{-1}(-\sqrt{3})$ . Then $\cot y_2 = -\sqrt{3}$ . The principal value range for $\cot^{-1}$ is $(0, \pi)$ . We know $\cot\left(\frac{\pi}{6}\right) = \sqrt{3}$ . So, $\cot y_2 = -\cot\left(\frac{\pi}{6}\right)$ . Using the identity $\cot(\pi - \theta) = -\cot \theta$ , we get: $\cot y_2 = \cot\left(\pi - \frac{\pi}{6}\right) = \cot\left(\frac{5\pi}{6}\right)$ . Since $\frac{5\pi}{6}$ is in the range $(0, \pi)$ , we have $\cot^{-1}(-\sqrt{3}) = \frac{5\pi}{6}$ .

Now, substitute these values into the expression: Expression =

\frac{\pi}{3} - \frac{5\pi}{6}

Find a common denominator: Expression =

\frac{2\pi}{6} - \frac{5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}

.

This corresponds to option (B).

Final Answer: (B)

-\frac{\pi}{2}

Q1Miscellaneous Exercise on Chapter 2

Find the value of the following: $\cos^{-1}\left(\cos \frac{13\pi}{6}\right)$

Solution

To Find: The value of

\cos^{-1}\left(\cos \frac{13\pi}{6}\right)

.

Solution: We know that

\cos^{-1}(\cos x) = x

only if

x

is in the principal value range of

\cos^{-1}

, which is

[0, \pi]

.

The angle given is

\frac{13\pi}{6}

, which is not in this range (since

\frac{13\pi}{6} > 2\pi > \pi

). We need to find an angle

\theta

in the range

[0, \pi]

such that

\cos \theta = \cos \frac{13\pi}{6}

.

We can rewrite the angle

\frac{13\pi}{6}

as:

\frac{13\pi}{6} = \frac{12\pi + \pi}{6} = 2\pi + \frac{\pi}{6}

.

Using the identity

\cos(2\pi + x) = \cos x

:

\cos\left(\frac{13\pi}{6}\right) = \cos\left(2\pi + \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right)

.

Now, the angle

\frac{\pi}{6}

is in the principal value range

[0, \pi]

.

Therefore, the expression becomes:

\cos^{-1}\left(\cos \frac{13\pi}{6}\right) = \cos^{-1}\left(\cos \frac{\pi}{6}\right) = \frac{\pi}{6}

.

Final Answer: The value is

\frac{\pi}{6}

.

Q2Miscellaneous Exercise on Chapter 2

Find the value of the following: $\tan^{-1}\left(\tan \frac{7\pi}{6}\right)$

Solution

To Find: The value of

\tan^{-1}\left(\tan \frac{7\pi}{6}\right)

.

Solution: We know that

\tan^{-1}(\tan x) = x

only if

x

is in the principal value range of

\tan^{-1}

, which is

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

.

The angle given is

\frac{7\pi}{6}

, which is not in this range. We need to find an angle

\theta

in the range

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

such that

\tan \theta = \tan \frac{7\pi}{6}

.

We can rewrite the angle

\frac{7\pi}{6}

as:

\frac{7\pi}{6} = \frac{6\pi + \pi}{6} = \pi + \frac{\pi}{6}

.

Using the identity

\tan(\pi + x) = \tan x

:

\tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right)

.

Now, the angle

\frac{\pi}{6}

is in the principal value range

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

.

Therefore, the expression becomes:

\tan^{-1}\left(\tan \frac{7\pi}{6}\right) = \tan^{-1}\left(\tan \frac{\pi}{6}\right) = \frac{\pi}{6}

.

Final Answer: The value is

\frac{\pi}{6}

.

Q3Miscellaneous Exercise on Chapter 2

Prove that $2 \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$

Solution

To Prove:

2 \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}

.

Proof: Let's start with the Left Hand Side (LHS). LHS =

2 \sin^{-1} \frac{3}{5}

.

First, convert

\sin^{-1} \frac{3}{5}

to

\tan^{-1}

. Let

\theta = \sin^{-1} \frac{3}{5}

, so

\sin \theta = \frac{3}{5}

. In a right-angled triangle, if opposite side = 3 and hypotenuse = 5, then the adjacent side is

\sqrt{5^2 - 3^2} = \sqrt{16} = 4

. Therefore,

\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}

. So,

\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}

.

Now, the LHS becomes: LHS =

2 \tan^{-1} \frac{3}{4}

.

We use the identity

2 \tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)

, which is valid for

|x| < 1

. Here

x = \frac{3}{4}

, and

|\frac{3}{4}| < 1

, so the identity can be applied.

LHS =

\tan^{-1}\left(\frac{2 \times \frac{3}{4}}{1 - (\frac{3}{4})^2}\right)

LHS =

\tan^{-1}\left(\frac{\frac{3}{2}}{1 - \frac{9}{16}}\right)

LHS =

\tan^{-1}\left(\frac{\frac{3}{2}}{\frac{16-9}{16}}\right)

LHS =

\tan^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)

LHS =

\tan^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)

LHS =

\tan^{-1}\left(\frac{3 \times 8}{7}\right)

LHS =

\tan^{-1}\left(\frac{24}{7}\right)

.

This is equal to the Right Hand Side (RHS).

Hence Proved.

Q4Miscellaneous Exercise on Chapter 2

Prove that $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$

Solution

To Prove:

\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}

.

Proof: Let's start with the Left Hand Side (LHS) and convert the

\sin^{-1}

terms to

\tan^{-1}

.

Convert $\sin^{-1} \frac{8}{17}$ : Let $\alpha = \sin^{-1} \frac{8}{17}$ , so $\sin \alpha = \frac{8}{17}$ . Adjacent side = $\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$ . So, $\tan \alpha = \frac{8}{15}$ . Thus, $\sin^{-1} \frac{8}{17} = \tan^{-1} \frac{8}{15}$ .
Convert $\sin^{-1} \frac{3}{5}$ : Let $\beta = \sin^{-1} \frac{3}{5}$ , so $\sin \beta = \frac{3}{5}$ . Adjacent side = $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$ . So, $\tan \beta = \frac{3}{4}$ . Thus, $\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$ .

Now the LHS becomes: LHS =

\tan^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}

.

We use the identity

\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)

. Here

x = \frac{8}{15}

and

y = \frac{3}{4}

. The product

xy = \frac{8}{15} \times \frac{3}{4} = \frac{24}{60} = \frac{2}{5} < 1

, so the formula is valid.

LHS =

\tan^{-1}\left(\frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \times \frac{3}{4}}\right)

LHS =

\tan^{-1}\left(\frac{\frac{32+45}{60}}{1 - \frac{24}{60}}\right)

LHS =

\tan^{-1}\left(\frac{\frac{77}{60}}{\frac{60-24}{60}}\right)

LHS =

\tan^{-1}\left(\frac{\frac{77}{60}}{\frac{36}{60}}\right)

LHS =

\tan^{-1}\left(\frac{77}{36}\right)

.

This is equal to the Right Hand Side (RHS).

Hence Proved.

Q5Miscellaneous Exercise on Chapter 2

Prove that $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$

Solution

To Prove:

\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}

.

Proof: Let

\alpha = \cos^{-1} \frac{4}{5}

and

\beta = \cos^{-1} \frac{12}{13}

. This implies

\cos \alpha = \frac{4}{5}

and

\cos \beta = \frac{12}{13}

. Since the inputs are positive,

\alpha

and

\beta

are in the first quadrant, so

\sin \alpha

and

\sin \beta

will be positive.

We can find

\sin \alpha

and

\sin \beta

using the identity

\sin^2 x + \cos^2 x = 1

.

\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}

.

\sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}

.

Now, we use the cosine addition formula:

\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta

.

\cos(\alpha + \beta) = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right)

\cos(\alpha + \beta) = \frac{48}{65} - \frac{15}{65}

\cos(\alpha + \beta) = \frac{33}{65}

This implies

\alpha + \beta = \cos^{-1} \frac{33}{65}

. Substituting back the values of

\alpha

and

\beta

:

\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}

.

This is the required result.

Hence Proved.

Q6Miscellaneous Exercise on Chapter 2

Prove that $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$

Solution

To Prove:

\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}

.

Proof: Let

\alpha = \cos^{-1} \frac{12}{13}

and

\beta = \sin^{-1} \frac{3}{5}

. This implies

\cos \alpha = \frac{12}{13}

and

\sin \beta = \frac{3}{5}

. Since the inputs are positive,

\alpha

and

\beta

are in the first quadrant.

We need to find

\sin \alpha

and

\cos \beta

.

\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}

.

\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}

.

Now, we use the sine addition formula:

\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta

.

\sin(\alpha + \beta) = \left(\frac{5}{13}\right)\left(\frac{4}{5}\right) + \left(\frac{12}{13}\right)\left(\frac{3}{5}\right)

\sin(\alpha + \beta) = \frac{20}{65} + \frac{36}{65}

\sin(\alpha + \beta) = \frac{56}{65}

This implies

\alpha + \beta = \sin^{-1} \frac{56}{65}

. Substituting back the values of

\alpha

and

\beta

:

\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}

.

This is the required result.

Hence Proved.

Q7Miscellaneous Exercise on Chapter 2

Prove that $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$

Solution

To Prove:

\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}

.

Proof: Let's start with the Right Hand Side (RHS) and convert the terms to

\tan^{-1}

. RHS =

\sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}

.

Convert $\sin^{-1} \frac{5}{13}$ : Let $\alpha = \sin^{-1} \frac{5}{13}$ , so $\sin \alpha = \frac{5}{13}$ . Adjacent side = $\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ . So, $\tan \alpha = \frac{5}{12}$ . Thus, $\sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}$ .
Convert $\cos^{-1} \frac{3}{5}$ : Let $\beta = \cos^{-1} \frac{3}{5}$ , so $\cos \beta = \frac{3}{5}$ . Opposite side = $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$ . So, $\tan \beta = \frac{4}{3}$ . Thus, $\cos^{-1} \frac{3}{5} = \tan^{-1} \frac{4}{3}$ .

Now the RHS becomes: RHS =

\tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}

.

We use the identity

\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)

. Here

x = \frac{5}{12}

and

y = \frac{4}{3}

. The product

xy = \frac{5}{12} \times \frac{4}{3} = \frac{20}{36} = \frac{5}{9} < 1

, so the formula is valid.

RHS =

\tan^{-1}\left(\frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \times \frac{4}{3}}\right)

RHS =

\tan^{-1}\left(\frac{\frac{5+16}{12}}{1 - \frac{20}{36}}\right)

RHS =

\tan^{-1}\left(\frac{\frac{21}{12}}{\frac{36-20}{36}}\right)

RHS =

\tan^{-1}\left(\frac{\frac{21}{12}}{\frac{16}{36}}\right)

RHS =

\tan^{-1}\left(\frac{21}{12} \times \frac{36}{16}\right)

RHS =

\tan^{-1}\left(\frac{21 \times 3}{16}\right)

RHS =

\tan^{-1}\left(\frac{63}{16}\right)

.

This is equal to the Left Hand Side (LHS).

Hence Proved.

Q8Miscellaneous Exercise on Chapter 2

Prove that $\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} \frac{1-x}{1+x}, x \in [0,1]$

Solution

To Prove:

\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} \frac{1-x}{1+x}

for

x \in [0,1]

.

Proof: Let's start with the Right Hand Side (RHS). RHS =

\frac{1}{2} \cos^{-1} \frac{1-x}{1+x}

.

Let

x = \tan^2 \theta

. Since

x \in [0,1]

,

\sqrt{x} = \tan \theta

where

\theta \in [0, \pi/4]

. So,

\theta = \tan^{-1} \sqrt{x}

.

Substitute

x = \tan^2 \theta

into the RHS: RHS =

\frac{1}{2} \cos^{-1} \left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right)

.

We use the double angle identity for cosine:

\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}

. RHS =

\frac{1}{2} \cos^{-1}(\cos 2\theta)

.

Now, we need to check the range of

2\theta

. Since

\theta \in [0, \pi/4]

, then

2\theta \in [0, \pi/2]

. This range is within the principal value range of

\cos^{-1}

, which is

[0, \pi]

. So,

\cos^{-1}(\cos 2\theta) = 2\theta

.

RHS =

\frac{1}{2} (2\theta) = \theta

.

Substitute back

\theta = \tan^{-1} \sqrt{x}

. RHS =

\tan^{-1} \sqrt{x}

.

This is equal to the Left Hand Side (LHS).

Hence Proved.

Q9Miscellaneous Exercise on Chapter 2

Prove that $\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2}, x \in \left(0, \frac{\pi}{4}\right)$

Solution

To Prove:

\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2}

for

x \in \left(0, \frac{\pi}{4}\right)

.

Proof: First, let's simplify the terms under the square roots. We use the identities

\sin x = 2\sin(x/2)\cos(x/2)

and

1 = \cos^2(x/2) + \sin^2(x/2)

.

1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2

.

1 - \sin x = \cos^2(x/2) + \sin^2(x/2) - 2\sin(x/2)\cos(x/2) = (\cos(x/2) - \sin(x/2))^2

.

So,

\sqrt{1+\sin x} = |\cos(x/2) + \sin(x/2)|

and

\sqrt{1-\sin x} = |\cos(x/2) - \sin(x/2)|

.

Given the range

x \in (0, \pi/4)

. This implies

x/2 \in (0, \pi/8)

. In this interval, both

\cos(x/2)

and

\sin(x/2)

are positive. Also,

\cos(x/2) > \sin(x/2)

. Therefore:

|\cos(x/2) + \sin(x/2)| = \cos(x/2) + \sin(x/2)

.

|\cos(x/2) - \sin(x/2)| = \cos(x/2) - \sin(x/2)

.

Now substitute these into the expression inside

\cot^{-1}

: Expression =

\frac{(\cos(x/2) + \sin(x/2)) + (\cos(x/2) - \sin(x/2))}{(\cos(x/2) + \sin(x/2)) - (\cos(x/2) - \sin(x/2))}

Expression =

\frac{2\cos(x/2)}{2\sin(x/2)} = \cot(x/2)

.

So, the Left Hand Side (LHS) becomes: LHS =

\cot^{-1}(\cot(x/2))

.

We need to check if

x/2

is in the principal value range of

\cot^{-1}

, which is

(0, \pi)

. Since

x/2 \in (0, \pi/8)

, it is within this range. Therefore,

\cot^{-1}(\cot(x/2)) = x/2

.

LHS =

x/2

, which is equal to the Right Hand Side (RHS).

Hence Proved.

Q10Miscellaneous Exercise on Chapter 2

Prove that $\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x, -\frac{1}{\sqrt{2}} \leq x \leq 1$ [Hint: Put $x=\cos 2\theta$ ]

Solution

To Prove:

\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x

.

Proof: Let's start with the Left Hand Side (LHS). Following the hint, we substitute

x = \cos 2\theta

. This implies

2\theta = \cos^{-1} x

, so

\theta = \frac{1}{2} \cos^{-1} x

.

Now, let's determine the range for

\theta

. Given

-\frac{1}{\sqrt{2}} \leq x \leq 1

.

-\frac{1}{\sqrt{2}} \leq \cos 2\theta \leq 1

. Applying

\cos^{-1}

(which reverses inequalities):

\cos^{-1}(1) \leq 2\theta \leq \cos^{-1}(-\frac{1}{\sqrt{2}})

.

0 \leq 2\theta \leq \frac{3\pi}{4}

. Dividing by 2:

0 \leq \theta \leq \frac{3\pi}{8}

.

Now, simplify the terms under the square root using

x = \cos 2\theta

and half-angle identities:

1+x = 1+\cos 2\theta = 2\cos^2\theta

. So

\sqrt{1+x} = \sqrt{2\cos^2\theta} = \sqrt{2}|\cos\theta|

.

1-x = 1-\cos 2\theta = 2\sin^2\theta

. So

\sqrt{1-x} = \sqrt{2\sin^2\theta} = \sqrt{2}|\sin\theta|

.

Since

\theta \in [0, 3\pi/8]

, both

\cos\theta

and

\sin\theta

are non-negative. So

|\cos\theta|=\cos\theta

and

|\sin\theta|=\sin\theta

.

\{\sqrt{1+x}} = \sqrt{2}\cos\theta

and

\{\sqrt{1-x}} = \sqrt{2}\sin\theta

.

Substitute these into the LHS: LHS =

\tan^{-1}\left(\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right)

LHS =

\tan^{-1}\left(\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}\right)

Divide numerator and denominator by

\cos\theta

(since

\theta \in [0, 3\pi/8]

,

\cos\theta \neq 0

): LHS =

\tan^{-1}\left(\frac{1 - \tan\theta}{1 + \tan\theta}\right)

Substitute

1 = \tan(\pi/4)

: LHS =

\tan^{-1}\left(\frac{\tan(\pi/4) - \tan\theta}{1 + \tan(\pi/4)\tan\theta}\right)

Using the identity

\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

: LHS =

\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right)

.

Now, check the range of

\frac{\pi}{4} - \theta

.

0 \leq \theta \leq \frac{3\pi}{8}

.

-\frac{3\pi}{8} \leq -\theta \leq 0

.

\frac{\pi}{4} - \frac{3\pi}{8} \leq \frac{\pi}{4} - \theta \leq \frac{\pi}{4} - 0

.

-\frac{\pi}{8} \leq \frac{\pi}{4} - \theta \leq \frac{\pi}{4}

. This range is within the principal value range of

\tan^{-1}

,

(-\pi/2, \pi/2)

. So,

\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta

.

LHS =

\frac{\pi}{4} - \theta

.

Finally, substitute back

\theta = \frac{1}{2} \cos^{-1} x

. LHS =

\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x

.

This is equal to the Right Hand Side (RHS).

Hence Proved.

Q11Miscellaneous Exercise on Chapter 2

Solve the following equations: $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)$

Solution

Given Equation:

2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)

.

To Solve: Find the value(s) of

x

.

Solution: We use the identity

2 \tan^{-1} y = \tan^{-1}\left(\frac{2y}{1-y^2}\right)

. Let

y = \cos x

. The Left Hand Side (LHS) becomes: LHS =

\tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right)

. Using the identity

1-\cos^2 x = \sin^2 x

: LHS =

\tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right)

.

Now, equate this to the Right Hand Side (RHS):

\tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right) = \tan^{-1}(2 \operatorname{cosec} x)

.

Applying

\tan

to both sides, we get:

\frac{2\cos x}{\sin^2 x} = 2 \operatorname{cosec} x

.

We know

\operatorname{cosec} x = \frac{1}{\sin x}

. For this to be defined,

\sin x \neq 0

.

\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}

.

Since

\sin x \neq 0

, we can multiply both sides by

\sin^2 x

:

2\cos x = 2\sin x

.

\cos x = \sin x

.

Divide both sides by

\cos x

(assuming

\cos x \neq 0

):

1 = \frac{\sin x}{\cos x}

.

\tan x = 1

.

The principal solution for this is

x = \frac{\pi}{4}

. If

\cos x = 0

, then

\sin x = \pm 1

. The equation

\cos x = \sin x

would not be satisfied, so our assumption was valid. The general solution is

x = n\pi + \frac{\pi}{4}

for any integer

n

. However, typically in these problems, we look for a solution in

[0, 2\pi)

or the principal solution.

Let's check the solution

x = \frac{\pi}{4}

. LHS =

2 \tan^{-1}(\cos \frac{\pi}{4}) = 2 \tan^{-1}(\frac{1}{\sqrt{2}})

. RHS =

\tan^{-1}(2 \operatorname{cosec} \frac{\pi}{4}) = \tan^{-1}(2\sqrt{2})

. Using

2\tan^{-1}y = \tan^{-1}(\frac{2y}{1-y^2})

, for LHS we have:

2 \tan^{-1}(\frac{1}{\sqrt{2}}) = \tan^{-1}\left(\frac{2(1/\sqrt{2})}{1-(1/\sqrt{2})^2}\right) = \tan^{-1}\left(\frac{\sqrt{2}}{1-1/2}\right) = \tan^{-1}\left(\frac{\sqrt{2}}{1/2}\right) = \tan^{-1}(2\sqrt{2})

. LHS = RHS. The solution is valid.

Final Answer:

x = \frac{\pi}{4}

.

Q12Miscellaneous Exercise on Chapter 2

Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x, (x>0)$

Solution

Given Equation:

\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x

, for

x>0

.

To Solve: Find the value(s) of

x

.

Solution: The term

\frac{1-x}{1+x}

can be simplified using a substitution. Let

x = \tan \theta

. Since

x > 0

, we have

\theta \in (0, \pi/2)

.

The Left Hand Side (LHS) can be rewritten using the identity

\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)

. Let

A=1

and

B=x

. Then

\tan^{-1} 1 - \tan^{-1} x = \tan^{-1} \frac{1-x}{1+x}

. Since

x>0

,

1 \cdot x > -1

, so this identity is valid.

So the equation becomes:

\tan^{-1} 1 - \tan^{-1} x = \frac{1}{2} \tan^{-1} x

.

We know

\tan^{-1} 1 = \frac{\pi}{4}

.

\frac{\pi}{4} - \tan^{-1} x = \frac{1}{2} \tan^{-1} x

.

Rearrange the terms to solve for

\tan^{-1} x

:

\frac{\pi}{4} = \frac{1}{2} \tan^{-1} x + \tan^{-1} x

.

\frac{\pi}{4} = \frac{3}{2} \tan^{-1} x

.

Multiply both sides by

\frac{2}{3}

:

\tan^{-1} x = \frac{\pi}{4} \times \frac{2}{3}

.

\tan^{-1} x = \frac{\pi}{6}

.

Now, apply

\tan

to both sides:

x = \tan\left(\frac{\pi}{6}\right)

.

x = \frac{1}{\sqrt{3}}

.

This solution satisfies the condition

x > 0

.

Final Answer:

x = \frac{1}{\sqrt{3}}

.

Q13Miscellaneous Exercise on Chapter 2

$\sin(\tan^{-1} x), |x| < 1$ is equal to

(A)

\frac{x}{\sqrt{1-x^2}}

(B)

\frac{1}{\sqrt{1-x^2}}

(C)

\frac{1}{\sqrt{1+x^2}}

(D)

\frac{x}{\sqrt{1+x^2}}

Solution

To Find: The value of

\sin(\tan^{-1} x)

for

|x| < 1

.

Solution: Let

\theta = \tan^{-1} x

. This implies

\tan \theta = x

. We can write this as

\tan \theta = \frac{x}{1}

.

We can visualize a right-angled triangle where

\theta

is an angle. Since

\tan \theta = \frac{\text{opposite}}{\text{adjacent}}

, we can set the opposite side to

x

and the adjacent side to 1.

By the Pythagorean theorem, the hypotenuse is: Hypotenuse =

\sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{x^2 + 1^2} = \sqrt{1+x^2}

.

Now we need to find

\sin(\tan^{-1} x)

, which is

\sin \theta

. From the triangle,

\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}

.

\sin \theta = \frac{x}{\sqrt{1+x^2}}

.

The condition

|x|<1

ensures that

\theta

is between

-\pi/4

and

\pi/4

, but the geometric construction is valid for all real

x

.

So,

\sin(\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}}

.

This corresponds to option (D).

Final Answer: (D)

\frac{x}{\sqrt{1+x^2}}

Q14Miscellaneous Exercise on Chapter 2

$\sin^{-1}(1-x) - 2\sin^{-1} x = \frac{\pi}{2}$ , then $x$ is equal to

(A)

0, \frac{1}{2}

(B)

1, \frac{1}{2}

(C)

0

(D)

\frac{1}{2}

Solution

Given Equation:

\sin^{-1}(1-x) - 2\sin^{-1} x = \frac{\pi}{2}

.

To Solve: Find the value(s) of

x

.

Solution: Rearrange the equation:

\sin^{-1}(1-x) = \frac{\pi}{2} + 2\sin^{-1} x

.

Apply the sine function to both sides:

1-x = \sin\left(\frac{\pi}{2} + 2\sin^{-1} x\right)

.

Using the identity

\sin(\frac{\pi}{2} + \theta) = \cos \theta

:

1-x = \cos(2\sin^{-1} x)

.

Let

\alpha = \sin^{-1} x

. The equation becomes:

1-x = \cos(2\alpha)

.

Using the double angle identity for cosine,

\cos(2\alpha) = 1 - 2\sin^2 \alpha

:

1-x = 1 - 2\sin^2(\sin^{-1} x)

.

Since

\sin(\sin^{-1} x) = x

, we have:

1-x = 1 - 2x^2

.

Now, solve the resulting quadratic equation for

x

:

-x = -2x^2

.

2x^2 - x = 0

.

x(2x - 1) = 0

.

This gives two possible solutions:

x=0

or

2x-1=0 \implies x=\frac{1}{2}

.

We must check both solutions in the original equation.

Check $x=0$ : LHS =

\sin^{-1}(1-0) - 2\sin^{-1}(0)

LHS =

\sin^{-1}(1) - 2(0)

LHS =

\frac{\pi}{2}

. RHS =

\frac{\pi}{2}

. So,

x=0

is a valid solution.

Check $x=\frac{1}{2}$ : LHS =

\sin^{-1}(1-\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2})

LHS =

\sin^{-1}(\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2})

LHS =

-\sin^{-1}(\frac{1}{2})

LHS =

-\frac{\pi}{6}

. RHS =

\frac{\pi}{2}

. Since LHS

\neq

RHS,

x=\frac{1}{2}

is not a solution.

The only valid solution is

x=0

.

This corresponds to option (C).

Final Answer: (C) 0

Inverse Trigonometric FunctionsClass 12 Mathematics NCERT Solutions

Find the principal values of the following: sin⁡−1(−12)\sin^{-1}\left(-\frac{1}{2}\right)sin−1(−21​)

Solution

Find the principal values of the following: cos⁡−1(32)\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)cos−1(23​​)

Solution

Find the principal values of the following: cosec⁡−1(2)\operatorname{cosec}^{-1}(2)cosec−1(2)

Solution

Find the principal values of the following: tan⁡−1(−3)\tan^{-1}(-\sqrt{3})tan−1(−3​)

Solution

Find the principal values of the following: cos⁡−1(−12)\cos^{-1}\left(-\frac{1}{2}\right)cos−1(−21​)

Solution

Find the principal values of the following: tan⁡−1(−1)\tan^{-1}(-1)tan−1(−1)

Solution

Find the principal values of the following: sec⁡−1(23)\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)sec−1(3​2​)

Solution

Find the principal values of the following: cot⁡−1(3)\cot^{-1}(\sqrt{3})cot−1(3​)

Solution

Find the principal values of the following: cos⁡−1(−12)\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)cos−1(−2​1​)

Solution

Find the principal values of the following: cosec⁡−1(−2)\operatorname{cosec}^{-1}(-\sqrt{2})cosec−1(−2​)

Solution

Find the values of the following: tan⁡−1(1)+cos⁡−1(−12)+sin⁡−1(−12)\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)tan−1(1)+cos−1(−21​)+sin−1(−21​)

Solution

Find the values of the following: cos⁡−1(12)+2sin⁡−1(12)\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)cos−1(21​)+2sin−1(21​)

Solution

If sin⁡−1x=y\sin^{-1} x = ysin−1x=y, then

Solution

tan⁡−13−sec⁡−1(−2)\tan^{-1} \sqrt{3} - \sec^{-1}(-2)tan−13​−sec−1(−2) is equal to

Solution

Prove the following: 3sin⁡−1x=sin⁡−1(3x−4x3),x∈[−12,12]3 \sin^{-1} x = \sin^{-1}(3x - 4x^3), x \in \left[-\frac{1}{2}, \frac{1}{2}\right]3sin−1x=sin−1(3x−4x3),x∈[−21​,21​]

Solution

Prove the following: 3cos⁡−1x=cos⁡−1(4x3−3x),x∈[12,1]3 \cos^{-1} x = \cos^{-1}(4x^3 - 3x), x \in \left[\frac{1}{2}, 1\right]3cos−1x=cos−1(4x3−3x),x∈[21​,1]

Solution

Write the following functions in the simplest form: tan⁡−11+x2−1x,x≠0\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}, x \neq 0tan−1x1+x2​−1​,x=0

Solution

Write the following functions in the simplest form: tan⁡−1(1−cos⁡x1+cos⁡x),0<x<π\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0 < x < \pitan−1(1+cosx1−cosx​​),0<x<π

Solution

Write the following functions in the simplest form: tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x),−π4<x<3π4\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right), \frac{-\pi}{4} < x < \frac{3\pi}{4}tan−1(cosx+sinxcosx−sinx​),4−π​<x<43π​

Solution

Write the following functions in the simplest form: tan⁡−1xa2−x2,∣x∣<a\tan^{-1} \frac{x}{\sqrt{a^2-x^2}}, |x| < atan−1a2−x2​x​,∣x∣<a

Solution

Write the following functions in the simplest form: tan⁡−1(3a2x−x3a3−3ax2),a>0;−a3<x<a3\tan^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}\right), a > 0; \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}tan−1(a3−3ax23a2x−x3​),a>0;3​−a​<x<3​a​

Solution

Find the values of each of the following: tan⁡−1[2cos⁡(2sin⁡−112)]\tan^{-1}\left[2 \cos \left(2 \sin^{-1} \frac{1}{2}\right)\right]tan−1[2cos(2sin−121​)]

Solution

Find the values of each of the following: tan⁡12[sin⁡−12x1+x2+cos⁡−11−y21+y2],∣x∣<1,y>0\tan \frac{1}{2}\left[\sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2}\right], |x| < 1, y > 0tan21​[sin−11+x22x​+cos−11+y21−y2​],∣x∣<1,y>0 and xy<1xy < 1xy<1

Solution

Find the values of each of the expressions in Exercises 16 to 18. sin⁡−1(sin⁡2π3)\sin^{-1}\left(\sin \frac{2\pi}{3}\right)sin−1(sin32π​)

Solution

tan⁡−1(tan⁡3π4)\tan^{-1}\left(\tan \frac{3\pi}{4}\right)tan−1(tan43π​)

Solution

tan⁡(sin⁡−135+cot⁡−132)\tan\left(\sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2}\right)tan(sin−153​+cot−123​)

Solution

cos⁡−1(cos⁡7π6)\cos^{-1}\left(\cos \frac{7\pi}{6}\right)cos−1(cos67π​) is equal to

Solution

sin⁡(π3−sin⁡−1(−12))\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)sin(3π​−sin−1(−21​)) is equal to

Solution

tan⁡−13−cot⁡−1(−3)\tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3})tan−13​−cot−1(−3​) is equal to

Solution

Find the value of the following: cos⁡−1(cos⁡13π6)\cos^{-1}\left(\cos \frac{13\pi}{6}\right)cos−1(cos613π​)

Solution

Find the value of the following: tan⁡−1(tan⁡7π6)\tan^{-1}\left(\tan \frac{7\pi}{6}\right)tan−1(tan67π​)

Solution

Prove that 2sin⁡−135=tan⁡−12472 \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}2sin−153​=tan−1724​

Solution

Prove that sin⁡−1817+sin⁡−135=tan⁡−17736\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}sin−1178​+sin−153​=tan−13677​

Solution

Prove that cos⁡−145+cos⁡−11213=cos⁡−13365\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}cos−154​+cos−11312​=cos−16533​

Solution

Prove that cos⁡−11213+sin⁡−135=sin⁡−15665\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}cos−11312​+sin−153​=sin−16556​

Solution

Prove that tan⁡−16316=sin⁡−1513+cos⁡−135\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}tan−11663​=sin−1135​+cos−153​

Solution

Prove that tan⁡−1x=12cos⁡−11−x1+x,x∈[0,1]\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} \frac{1-x}{1+x}, x \in [0,1]tan−1x​=21​cos−11+x1−x​,x∈[0,1]

Solution

Solution

Solution

Solve the following equations: 2tan⁡−1(cos⁡x)=tan⁡−1(2cosec⁡x)2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)2tan−1(cosx)=tan−1(2cosecx)

Solution

Solve the following equations: tan⁡−11−x1+x=12tan⁡−1x,(x>0)\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x, (x>0)tan−11+x1−x​=21​tan−1x,(x>0)

Find the principal values of the following: $\sin^{-1}\left(-\frac{1}{2}\right)$

Find the principal values of the following: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Find the principal values of the following: $\operatorname{cosec}^{-1}(2)$

Find the principal values of the following: $\tan^{-1}(-\sqrt{3})$

Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{2}\right)$

Find the principal values of the following: $\tan^{-1}(-1)$

Find the principal values of the following: $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Find the principal values of the following: $\cot^{-1}(\sqrt{3})$

Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$

Find the principal values of the following: $\operatorname{cosec}^{-1}(-\sqrt{2})$

Find the values of the following: $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$

Find the values of the following: $\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)$

If $\sin^{-1} x = y$ , then

$\tan^{-1} \sqrt{3} - \sec^{-1}(-2)$ is equal to

Prove the following: $3 \sin^{-1} x = \sin^{-1}(3x - 4x^3), x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$

Prove the following: $3 \cos^{-1} x = \cos^{-1}(4x^3 - 3x), x \in \left[\frac{1}{2}, 1\right]$

Write the following functions in the simplest form: $\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}, x \neq 0$

Write the following functions in the simplest form: $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0 < x < \pi$

Write the following functions in the simplest form: $\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right), \frac{-\pi}{4} < x < \frac{3\pi}{4}$

Write the following functions in the simplest form: $\tan^{-1} \frac{x}{\sqrt{a^2-x^2}}, |x| < a$

Write the following functions in the simplest form: $\tan^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}\right), a > 0; \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$

Find the values of each of the following: $\tan^{-1}\left[2 \cos \left(2 \sin^{-1} \frac{1}{2}\right)\right]$

Find the values of each of the following: $\tan \frac{1}{2}\left[\sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2}\right], |x| < 1, y > 0$ and $xy < 1$

Find the values of each of the expressions in Exercises 16 to 18. $\sin^{-1}\left(\sin \frac{2\pi}{3}\right)$

$\tan^{-1}\left(\tan \frac{3\pi}{4}\right)$

$\tan\left(\sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2}\right)$

$\cos^{-1}\left(\cos \frac{7\pi}{6}\right)$ is equal to

$\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to

$\tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3})$ is equal to

Find the value of the following: $\cos^{-1}\left(\cos \frac{13\pi}{6}\right)$

Find the value of the following: $\tan^{-1}\left(\tan \frac{7\pi}{6}\right)$

Prove that $2 \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$

Prove that $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$

Prove that $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$

Prove that $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$

Prove that $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$

Prove that $\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} \frac{1-x}{1+x}, x \in [0,1]$

Solve the following equations: $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)$

Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x, (x>0)$

$\sin(\tan^{-1} x), |x| < 1$ is equal to

$\sin^{-1}(1-x) - 2\sin^{-1} x = \frac{\pi}{2}$ , then $x$ is equal to