Question 1

Q1: Find the principal values of the following: $\sin^{-1}\left(-\frac{1}{2}\right)$

Accepted Answer

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 = -\f...

Question 2

Q10: Find the principal values of the following: $\operatorname{cosec}^{-1}(-\sqrt{2})$

Accepted Answer

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}{...

Question 3

Q11: Find the values of the following: $	an^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}ight) + \sin^{-1}\left(-\frac{1}{2}ight)$

Accepted Answer

To Find: The value of the expression $	an^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}ight) + \sin^{-1}\left(-\frac{1}{2}ight)$. Solution: We will find the principal value of each term separately. 1.  For $	an^{-1}(1)$:     Let $y_1 = 	an^{-1}(1)$. Then $	an y_1 = 1 = 	an\left(\frac{\pi}{4}ight)...

Question 4

Q12: Find the values of the following: $\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)$

Accepted Answer

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. 1.  For $\cos^{-1}\left(\frac{1}{2}\right)$:     Let $y_1 = \cos^{-1}\left(\frac{1}{2}\right)$. Then $\cos y_1 = \frac{1}...

Question 5

Q13: 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{\p...

Accepted Answer

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 \f...

Question 6

Q14: $	an^{-1} \sqrt{3} - \sec^{-1}(-2)$ is equal to (A) $\pi$ (B) $-\frac{\pi}{3}$ (C) $\frac{\pi}{3}$ (D) $\frac{2\pi}{3}$

Accepted Answer

To Find: The value of the expression $	an^{-1} \sqrt{3} - \sec^{-1}(-2)$. Solution: We will find the principal value of each term separately. 1.  For $	an^{-1} \sqrt{3}$:     Let $y_1 = 	an^{-1} \sqrt{3}$. Then $	an y_1 = \sqrt{3} = 	an\left(\frac{\pi}{3}ight)$.     The principal value range...

Question 7

Q2: Find the principal values of the following: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Accepted Answer

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

Question 8

Q3: Find the principal values of the following: $\operatorname{cosec}^{-1}(2)$

Accepted Answer

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 h...

Question 9

Q4: Find the principal values of the following: $	an^{-1}(-\sqrt{3})$

Accepted Answer

To Find: The principal value of $	an^{-1}(-\sqrt{3})$. Solution: Let $y = 	an^{-1}(-\sqrt{3})$. This means $	an y = -\sqrt{3}$. We know that the range of the principal value branch of $	an^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$. We have $	an y = -\sqrt{3}$. Since $	an\left(\frac...

Question 10

Q5: Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{2}\right)$

Accepted Answer

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...

Question 11

Q6: Find the principal values of the following: $	an^{-1}(-1)$

Accepted Answer

To Find: The principal value of $	an^{-1}(-1)$. Solution: Let $y = 	an^{-1}(-1)$. This means $	an y = -1$. We know that the range of the principal value branch of $	an^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$. We have $	an y = -1$. Since $	an\left(\frac{\pi}{4}ight) = 1$, we can...

Question 12

Q7: Find the principal values of the following: $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Accepted Answer

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 hav...

Question 13

Q8: Find the principal values of the following: $\cot^{-1}(\sqrt{3})$

Accepted Answer

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

Question 14

Q9: Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$

Accepted Answer

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{...

Question 15

Q1: Prove the following: $3 \sin^{-1} x = \sin^{-1}(3x - 4x^3), x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$

Accepted Answer

To Prove: $3 \sin^{-1} x = \sin^{-1}(3x - 4x^3)$ for $x \in \left[-\frac{1}{2}, \frac{1}{2}ight]$. Proof: Let $\sin^{-1} x = 	heta$. This implies $x = \sin 	heta$. Now, we work with the Right Hand Side (RHS) of the equation: RHS = $\sin^{-1}(3x - 4x^3)$ Substitute $x = \sin 	heta$ into the expr...

Question 16

Q10: Find the values of each of the expressions in Exercises 16 to 18. $\sin^{-1}\left(\sin \frac{2\pi}{3}\right)$

Accepted Answer

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 nee...

Question 17

Q11: $	an^{-1}\left(	an \frac{3\pi}{4}ight)$

Accepted Answer

To Find: The value of $	an^{-1}\left(	an \frac{3\pi}{4}ight)$. Solution: We know that $	an^{-1}(	an x) = x$ only if $x$ is in the principal value range of $	an^{-1}$, which is $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$. The angle given is $\frac{3\pi}{4}$, which is not in this range. We nee...

Question 18

Q12: $	an\left(\sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2}ight)$

Accepted Answer

To Find: The value of $	an\left(\sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2}ight)$. Solution: To simplify the expression, we convert the inverse functions inside the parentheses to $	an^{-1}$. 1.  Convert $\sin^{-1} \frac{3}{5}$ to $	an^{-1}$:     Let $\alpha = \sin^{-1} \frac{3}{5}$. This mea...

Question 19

Q13: $\cos^{-1}\left(\cos \frac{7\pi}{6}\right)$ is equal to (A) $\frac{7\pi}{6}$ (B) $\frac{5\pi}{6}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{6}$

Accepted Answer

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 t...

Question 20

Q14: $\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to (A) $\frac{1}{2}$ (B) $\frac{1}{3}$ (C) $\frac{1}{4}$ (D) 1

Accepted Answer

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...

NCERT Solutions