Q3: If sin A=3/4, calculate cos A and tan A.

Given: sin A = 3/4 To Find: cos A and tan A. Solution: We know that for an acute angle A in a right-angled triangle, sin A = Opposite Side/Hypotenuse = 3/4 Let the opposite side be 3k and the hypotenuse be 4k, where k is a positive constant. Using the Pythagoras theorem, (Hypotenuse)^2 =…

Q4: Given 15 cot A=8, find sin A and sec A.

Given: 15 cot A = 8, which means cot A = 8/15. To Find: sin A and sec A. Solution: We know that for an acute angle A in a right-angled triangle, cot A = Adjacent Side/Opposite Side = 8/15 Let the adjacent side be 8k and the opposite side be 15k, where k is a positive constant. Using the Pythagoras…

Q5: Given sec θ=13/12, calculate all other trigonometric ratios.

Given: sec θ = 13/12 To Find: All other trigonometric ratios ( sin θ, cos θ, tan θ, cot θ, csc θ ) Solution: We know that for an acute angle θ in a right-angled triangle, sec θ = Hypotenuse/Adjacent Side = 13/12 Let the hypotenuse be 13k and the adjacent side be 12k, where k is a positive constant.…

Q7 (i): (1+sin θ)(1-sin θ)/(1+cos θ)(1-cos θ),

Given: cot θ = 7/8 To Evaluate: (i) (1+sin θ)(1-sin θ)/(1+cos θ)(1-cos θ) (ii) cot^2 θ Solution: (i) Evaluate (1+sin θ)(1-sin θ)/(1+cos θ)(1-cos θ) We can simplify the expression using the algebraic identity (a+b)(a-b) = a^2 - b^2. Numerator: (1+sin θ)(1-sin θ) = 1^2 - sin^2 θ = 1 - sin^2 θ.…

Q8: If 3 cot A=4, check whether frac1-tan ^2 ~A1+tan ^2 ~A=cos ^2 ~A-sin ^2 ~A or not.

Given: 3 cot A = 4, which implies cot A = 4/3. To Verify: Whether 1-tan^2 A/1+tan^2 A = cos^2 A - sin^2 A. Solution: From the given cot A = 4/3, we can find tan A. tan A = 1/cot A = 1/4/3 = 3/4 Now, let's find the values of sin A and cos A. We know cot A = Adjacent/Opposite = 4/3. Let Adjacent side…

Q9 (i): sin A cos C+cos A sin C

Given: In triangle ABC, angle B = 90^°. tan A = frac1√(3). To Find: (i) sin A cos C + cos A sin C (ii) cos A cos C - sin A sin C Solution: Given tan A = frac1√(3). We know tan A = Opposite/Adjacent = BC/AB. So, let BC = k and AB = k√(3) for some positive constant k. By Pythagoras theorem: AC =…

Q9 (ii): cos A cos C-sin A sin C

Given: In triangle ABC, angle B = 90^°. tan A = frac1√(3). To Find: (i) sin A cos C + cos A sin C (ii) cos A cos C - sin A sin C Solution: Given tan A = frac1√(3). We know tan A = Opposite/Adjacent = BC/AB. So, let BC = k and AB = k√(3) for some positive constant k. By Pythagoras theorem: AC =…

Q1 (i): sin 60^° cos 30^°+sin 30^° cos 60^°

Solution: (i) sin 60^° cos 30^°+sin 30^° cos 60^° We know the values: sin 60^° = frac√(3)2, cos 30^° = frac√(3)2, sin 30^° = 1/2, cos 60^° = 1/2. = left(frac√(3)2right) left(frac√(3)2right) + left(1/2right) left(1/2right) = 3/4 + 1/4 = 3+1/4 = 4/4 = 1 Final Answer (i): 1 (ii) 2 tan ^2 45^°+cos ^2…

Q1 (ii): 2 tan ^2 45^°+cos ^2 30^°-sin ^2 60^°

Solution: (i) sin 60^° cos 30^°+sin 30^° cos 60^° We know the values: sin 60^° = frac√(3)2, cos 30^° = frac√(3)2, sin 30^° = 1/2, cos 60^° = 1/2. = left(frac√(3)2right) left(frac√(3)2right) + left(1/2right) left(1/2right) = 3/4 + 1/4 = 3+1/4 = 4/4 = 1 Final Answer (i): 1 (ii) 2 tan ^2 45^°+cos ^2…

Introduction to Trigonometry Class 10 Mathematics NCERT Solutions

Q1EXERCISE 8.1

In $\triangle \mathrm{ABC}$ , right-angled at $\mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}, \mathrm{BC}=7 \mathrm{~cm}$ . Determine :

(i)

$\sin \mathrm{A}, \cos \mathrm{A}$

(ii)

$\sin \mathrm{C}, \cos \mathrm{C}$

Solution

Given: A right-angled triangle ABC, with the right angle at B. Side AB = 24 cm. Side BC = 7 cm.

To Find:

(i)

\sin A, \cos A

(ii)

\sin C, \cos C

Solution: First, we need to find the length of the hypotenuse AC using the Pythagoras theorem. In

\triangle ABC

,

AC^2 = AB^2 + BC^2

AC^2 = (24)^2 + (7)^2

AC^2 = 576 + 49

AC^2 = 625

AC = \sqrt{625} = 25 \text{ cm}

Now we can determine the trigonometric ratios.

(i) For angle A: Side opposite to angle A = BC = 7 cm. Side adjacent to angle A = AB = 24 cm. Hypotenuse = AC = 25 cm.

\sin A = \frac{\text{Side opposite to angle A}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}

\cos A = \frac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}

(ii) For angle C: Side opposite to angle C = AB = 24 cm. Side adjacent to angle C = BC = 7 cm. Hypotenuse = AC = 25 cm.

\sin C = \frac{\text{Side opposite to angle C}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}

\cos C = \frac{\text{Side adjacent to angle C}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}

Final Answer:

(i)

\sin A = \frac{7}{25}

,

\cos A = \frac{24}{25}

(ii)

\sin C = \frac{24}{25}

,

\cos C = \frac{7}{25}

Q2EXERCISE 8.1

In Fig. 8.13, find $\tan \mathrm{P}-\cot \mathrm{R}$ .

Solution

Given: A right-angled triangle PQR, with the right angle at Q. Side PQ = 12 cm. Side PR (Hypotenuse) = 13 cm.

To Find: The value of

\tan P - \cot R

.

Solution: First, we find the length of the side QR using the Pythagoras theorem. In

\triangle PQR

,

PR^2 = PQ^2 + QR^2

(13)^2 = (12)^2 + QR^2

169 = 144 + QR^2

QR^2 = 169 - 144 = 25

QR = \sqrt{25} = 5 \text{ cm}

Now, we find the values of

\tan P

and

\cot R

.

For angle P: Side opposite to angle P = QR = 5 cm. Side adjacent to angle P = PQ = 12 cm.

\tan P = \frac{\text{Side opposite to angle P}}{\text{Side adjacent to angle P}} = \frac{QR}{PQ} = \frac{5}{12}

For angle R: Side opposite to angle R = PQ = 12 cm. Side adjacent to angle R = QR = 5 cm.

\cot R = \frac{\text{Side adjacent to angle R}}{\text{Side opposite to angle R}} = \frac{QR}{PQ} = \frac{5}{12}

Now, we calculate

\tan P - \cot R

.

\tan P - \cot R = \frac{5}{12} - \frac{5}{12} = 0

Final Answer:

\tan P - \cot R = 0

Q3EXERCISE 8.1

If $\sin \mathrm{A}=\frac{3}{4}$ , calculate $\cos \mathrm{A}$ and $\tan \mathrm{A}$ .

Solution

Given:

\sin A = \frac{3}{4}

To Find:

\cos A

and

\tan A

.

Solution: We know that for an acute angle A in a right-angled triangle,

\sin A = \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{3}{4}

Let the opposite side be

3k

and the hypotenuse be

4k

, where

k

is a positive constant.

Using the Pythagoras theorem,

(\text{Hypotenuse})^2 = (\text{Opposite})^2 + (\text{Adjacent})^2

.

(4k)^2 = (3k)^2 + (\text{Adjacent})^2

16k^2 = 9k^2 + (\text{Adjacent})^2

(\text{Adjacent})^2 = 16k^2 - 9k^2 = 7k^2

\text{Adjacent} = \sqrt{7k^2} = k\sqrt{7}

Now we can find

\cos A

and

\tan A

.

\cos A = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{k\sqrt{7}}{4k} = \frac{\sqrt{7}}{4}

\tan A = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{3k}{k\sqrt{7}} = \frac{3}{\sqrt{7}}

Final Answer:

\cos A = \frac{\sqrt{7}}{4}

and

\tan A = \frac{3}{\sqrt{7}}

Q4EXERCISE 8.1

Given $15 \cot \mathrm{A}=8$ , find $\sin \mathrm{A}$ and $\sec \mathrm{A}$ .

Solution

Given:

15 \cot A = 8

, which means

\cot A = \frac{8}{15}

.

To Find:

\sin A

and

\sec A

.

Solution: We know that for an acute angle A in a right-angled triangle,

\cot A = \frac{\text{Adjacent Side}}{\text{Opposite Side}} = \frac{8}{15}

Let the adjacent side be

8k

and the opposite side be

15k

, where

k

is a positive constant.

Using the Pythagoras theorem,

(\text{Hypotenuse})^2 = (\text{Opposite})^2 + (\text{Adjacent})^2

.

(\text{Hypotenuse})^2 = (15k)^2 + (8k)^2

(\text{Hypotenuse})^2 = 225k^2 + 64k^2 = 289k^2

\text{Hypotenuse} = \sqrt{289k^2} = 17k

Now we can find

\sin A

and

\sec A

.

\sin A = \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{15k}{17k} = \frac{15}{17}

\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent Side}} = \frac{17k}{8k} = \frac{17}{8}

Final Answer:

\sin A = \frac{15}{17}

and

\sec A = \frac{17}{8}

Q5EXERCISE 8.1

Given $\sec \theta=\frac{13}{12}$ , calculate all other trigonometric ratios.

Solution

Given:

\sec \theta = \frac{13}{12}

To Find: All other trigonometric ratios (

\sin \theta, \cos \theta, \tan \theta, \cot \theta, \csc \theta

)

Solution: We know that for an acute angle

\theta

in a right-angled triangle,

\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent Side}} = \frac{13}{12}

Let the hypotenuse be

13k

and the adjacent side be

12k

, where

k

is a positive constant.

Using the Pythagoras theorem,

(\text{Hypotenuse})^2 = (\text{Opposite})^2 + (\text{Adjacent})^2

.

(13k)^2 = (\text{Opposite})^2 + (12k)^2

169k^2 = (\text{Opposite})^2 + 144k^2

(\text{Opposite})^2 = 169k^2 - 144k^2 = 25k^2

\text{Opposite} = \sqrt{25k^2} = 5k

Now we can find all other trigonometric ratios.

\sin \theta = \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{5k}{13k} = \frac{5}{13}

\cos \theta = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{12k}{13k} = \frac{12}{13}

\tan \theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{5k}{12k} = \frac{5}{12}

\csc \theta = \frac{1}{\sin \theta} = \frac{13}{5}

\cot \theta = \frac{1}{\tan \theta} = \frac{12}{5}

Final Answer:

\sin \theta = \frac{5}{13}

,

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

,

\tan \theta = \frac{5}{12}

,

\csc \theta = \frac{13}{5}

,

\cot \theta = \frac{12}{5}

Q6EXERCISE 8.1

If $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles such that $\cos \mathrm{A}=\cos \mathrm{B}$ , then show that $\angle \mathrm{A}=\angle \mathrm{B}$ .

Solution

Given:

\angle A

and

\angle B

are acute angles.

\cos A = \cos B

.

To Prove:

\angle A = \angle B

.

Proof: Let us consider two right-angled triangles,

\triangle APQ

and

\triangle BRS

, such that

\angle P = 90^\circ

and

\angle R = 90^\circ

.

From

\triangle APQ

, we have:

\cos A = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{AP}{AQ}

From

\triangle BRS

, we have:

\cos B = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{BR}{BS}

Since it is given that

\cos A = \cos B

,

\frac{AP}{AQ} = \frac{BR}{BS}

Let this ratio be equal to a constant

k

. So,

\frac{AP}{BR} = \frac{AQ}{BS} = k

This implies

AP = k(BR)

and

AQ = k(BS)

.

Now, let us find the ratio of the third sides, PQ and RS, using the Pythagoras theorem. In

\triangle APQ

:

PQ = \sqrt{AQ^2 - AP^2} = \sqrt{(k \cdot BS)^2 - (k \cdot BR)^2} = \sqrt{k^2(BS^2 - BR^2)} = k \sqrt{BS^2 - BR^2}

In

\triangle BRS

:

RS = \sqrt{BS^2 - BR^2}

Now, let's find the ratio

\frac{PQ}{RS}

:

\frac{PQ}{RS} = \frac{k \sqrt{BS^2 - BR^2}}{\sqrt{BS^2 - BR^2}} = k

So, we have the ratios of all corresponding sides equal:

\frac{AP}{BR} = \frac{AQ}{BS} = \frac{PQ}{RS} = k

By the SSS (Side-Side-Side) similarity criterion,

\triangle APQ \sim \triangle BRS

. Since the triangles are similar, their corresponding angles must be equal. Therefore,

\angle A = \angle B

.

Hence Proved.

Q7EXERCISE 8.1

If $\cot \theta=\frac{7}{8}$ , evaluate :

(i)

$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ ,

(ii)

$\cot ^{2} \theta$

Solution

Given:

\cot \theta = \frac{7}{8}

To Evaluate:

(i)

\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}

(ii)

\cot^2 \theta

Solution:

(i) Evaluate $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$

We can simplify the expression using the algebraic identity

(a+b)(a-b) = a^2 - b^2

. Numerator:

(1+\sin \theta)(1-\sin \theta) = 1^2 - \sin^2 \theta = 1 - \sin^2 \theta

. Denominator:

(1+\cos \theta)(1-\cos \theta) = 1^2 - \cos^2 \theta = 1 - \cos^2 \theta

.

Using the trigonometric identity

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

, we have:

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

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

So, the expression becomes:

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

Since

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

, the expression is equal to

\cot^2 \theta

.

Given

\cot \theta = \frac{7}{8}

,

\cot^2 \theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}

(ii) Evaluate $\cot^2 \theta$

This is a direct calculation from the given value. Given

\cot \theta = \frac{7}{8}

,

\cot^2 \theta = (\cot \theta)^2 = \left(\frac{7}{8}\right)^2 = \frac{49}{64}

Final Answer:

(i)

\frac{49}{64}

(ii)

\frac{49}{64}

Q8EXERCISE 8.1

If $3 \cot \mathrm{A}=4$ , check whether $\frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}$ or not.

Solution

Given:

3 \cot A = 4

, which implies

\cot A = \frac{4}{3}

.

To Verify: Whether

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

.

Solution: From the given

\cot A = \frac{4}{3}

, we can find

\tan A

.

\tan A = \frac{1}{\cot A} = \frac{1}{4/3} = \frac{3}{4}

Now, let's find the values of

\sin A

and

\cos A

. We know

\cot A = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{4}{3}

. Let Adjacent side =

4k

and Opposite side =

3k

. By Pythagoras theorem, Hypotenuse =

\sqrt{(4k)^2 + (3k)^2} = \sqrt{16k^2 + 9k^2} = \sqrt{25k^2} = 5k

.

So,

\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3k}{5k} = \frac{3}{5}

\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4k}{5k} = \frac{4}{5}

Now, we evaluate both sides of the equation separately.

LHS (Left Hand Side):

\frac{1-\tan^2 A}{1+\tan^2 A} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{(16-9)/16}{(16+9)/16} = \frac{7/16}{25/16} = \frac{7}{25}

RHS (Right Hand Side):

\cos^2 A - \sin^2 A = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{16-9}{25} = \frac{7}{25}

Since LHS = RHS, the statement is true.

Final Answer: Yes,

\frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}

is true.

Q9EXERCISE 8.1

In triangle ABC , right-angled at B , if $\tan \mathrm{A}=\frac{1}{\sqrt{3}}$ , find the value of:

(i)

$\sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C}$

(ii)

$\cos \mathrm{A} \cos \mathrm{C}-\sin \mathrm{A} \sin \mathrm{C}$

Solution

Given: In

\triangle ABC

,

\angle B = 90^\circ

.

\tan A = \frac{1}{\sqrt{3}}

.

To Find:

(i)

\sin A \cos C + \cos A \sin C

(ii)

\cos A \cos C - \sin A \sin C

Solution: Given

\tan A = \frac{1}{\sqrt{3}}

. We know

\tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB}

. So, let

BC = k

and

AB = k\sqrt{3}

for some positive constant

k

.

By Pythagoras theorem:

AC = \sqrt{AB^2 + BC^2} = \sqrt{(k\sqrt{3})^2 + (k)^2} = \sqrt{3k^2 + k^2} = \sqrt{4k^2} = 2k

Now, we find the trigonometric ratios for angles A and C. For angle A:

\sin A = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}

\cos A = \frac{AB}{AC} = \frac{k\sqrt{3}}{2k} = \frac{\sqrt{3}}{2}

For angle C: Side opposite to C is AB, and side adjacent to C is BC.

\sin C = \frac{AB}{AC} = \frac{k\sqrt{3}}{2k} = \frac{\sqrt{3}}{2}

\cos C = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}

Now we can evaluate the expressions.

(i) $\sin A \cos C + \cos A \sin C$

= \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right)

= \frac{1}{4} + \frac{3}{4} = \frac{1+3}{4} = \frac{4}{4} = 1

(ii) $\cos A \cos C - \sin A \sin C$

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

= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0

Final Answer:

(i)

1

(ii)

0

Q10EXERCISE 8.1

In $\triangle \mathrm{PQR}$ , right-angled at $\mathrm{Q}, \mathrm{PR}+\mathrm{QR}=25 \mathrm{~cm}$ and $\mathrm{PQ}=5 \mathrm{~cm}$ . Determine the values of $\sin \mathrm{P}, \cos \mathrm{P}$ and tan P .

Solution

Given: In

\triangle PQR

,

\angle Q = 90^\circ

.

PQ = 5

cm.

PR + QR = 25

cm.

To Find: The values of

\sin P, \cos P

, and

\tan P

.

Solution: From the given condition,

PR + QR = 25

, we can write

PR = 25 - QR

.

Using the Pythagoras theorem in

\triangle PQR

:

PR^2 = PQ^2 + QR^2

Substitute the values of PR and PQ:

(25 - QR)^2 = (5)^2 + QR^2

25^2 - 2(25)(QR) + QR^2 = 25 + QR^2

625 - 50 \cdot QR + QR^2 = 25 + QR^2

625 - 50 \cdot QR = 25

625 - 25 = 50 \cdot QR

600 = 50 \cdot QR

QR = \frac{600}{50} = 12 \text{ cm}

Now, find PR:

PR = 25 - QR = 25 - 12 = 13 \text{ cm}

So, the sides of the triangle are:

PQ = 5

cm (Adjacent to P)

QR = 12

cm (Opposite to P)

PR = 13

cm (Hypotenuse)

Now, we can find the required trigonometric ratios for angle P.

\sin P = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{12}{13}

\cos P = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{5}{13}

\tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ} = \frac{12}{5}

Final Answer:

\sin P = \frac{12}{13}

,

\cos P = \frac{5}{13}

, and

\tan P = \frac{12}{5}

.

Q11EXERCISE 8.1

State whether the following are true or false. Justify your answer.

(i)

The value of $\tan \mathrm{A}$ is always less than 1 .

(ii)

$\sec \mathrm{A}=\frac{12}{5}$ for some value of angle A .

(iii)

$\cos \mathrm{A}$ is the abbreviation used for the cosecant of angle A .

(iv)

$\cot \mathrm{A}$ is the product of $\cot$ and A .

(v)

$\sin \theta=\frac{4}{3}$ for some angle $\theta$ .

Solution

(i) The value of $\tan \mathrm{A}$ is always less than 1. Answer: False. Reason:

\tan A = \frac{\text{Opposite Side}}{\text{Adjacent Side}}

. The opposite side can be greater than, equal to, or less than the adjacent side. For example, in a right triangle with sides 3, 4, 5, if the opposite side is 4 and the adjacent side is 3, then

\tan A = \frac{4}{3} > 1

. Therefore, the value of

\tan A

is not always less than 1.

(ii) $\sec \mathrm{A}=\frac{12}{5}$ for some value of angle A. Answer: True. Reason:

\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent Side}}

. The hypotenuse is always the longest side in a right-angled triangle, so its length is always greater than or equal to the length of the adjacent side. The ratio

\frac{12}{5}

is greater than 1, which is possible for

\sec A

. Therefore,

\sec A = \frac{12}{5}

is possible for some value of A.

(iii) $\cos \mathrm{A}$ is the abbreviation used for the cosecant of angle A. Answer: False. Reason:

\cos A

is the abbreviation for 'cosine of angle A'. The abbreviation for 'cosecant of angle A' is

\csc A

or

\text{cosec } A

.

(iv) $\cot \mathrm{A}$ is the product of $\cot$ and A. Answer: False. Reason:

\cot A

is a single term representing the cotangent of the angle A. 'cot' separated from 'A' has no meaning. It is a function of the angle A, not a product.

(v) $\sin \theta=\frac{4}{3}$ for some angle $\theta$ . Answer: False. Reason:

\sin \theta = \frac{\text{Opposite Side}}{\text{Hypotenuse}}

. In a right-angled triangle, the hypotenuse is always the longest side. Therefore, the opposite side can never be greater than the hypotenuse. This means the value of

\sin \theta

cannot be greater than 1. Since

\frac{4}{3} > 1

, this value is not possible for

\sin \theta

.

Q1EXERCISE 8.2

Evaluate the following:

(i)

$\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$

(ii)

$2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

(iii)

$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$

(iv)

$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$

(v)

$\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$

Solution

Solution:

(i) $\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$ We know the values:

\sin 60^{\circ} = \frac{\sqrt{3}}{2}

,

\cos 30^{\circ} = \frac{\sqrt{3}}{2}

,

\sin 30^{\circ} = \frac{1}{2}

,

\cos 60^{\circ} = \frac{1}{2}

.

= \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)

= \frac{3}{4} + \frac{1}{4} = \frac{3+1}{4} = \frac{4}{4} = 1

Final Answer (i): 1

(ii) $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$ We know the values:

\tan 45^{\circ} = 1

,

\cos 30^{\circ} = \frac{\sqrt{3}}{2}

,

\sin 60^{\circ} = \frac{\sqrt{3}}{2}

.

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

= 2(1) + \frac{3}{4} - \frac{3}{4} = 2

Final Answer (ii): 2

(iii) $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$ We know the values:

\cos 45^{\circ} = \frac{1}{\sqrt{2}}

,

\sec 30^{\circ} = \frac{2}{\sqrt{3}}

,

\csc 30^{\circ} = 2

.

= \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2} = \frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}}

= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2+2\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{2}(2+2\sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})}

To rationalize the denominator, multiply numerator and denominator by

\sqrt{2}(\sqrt{3}-1)

:

= \frac{\sqrt{3} \times \sqrt{2}(\sqrt{3}-1)}{2\sqrt{2}(1+\sqrt{3}) \times \sqrt{2}(\sqrt{3}-1)} = \frac{\sqrt{6}(\sqrt{3}-1)}{4(3-1)} = \frac{3\sqrt{2}-\sqrt{6}}{4(2)} = \frac{3\sqrt{2}-\sqrt{6}}{8}

Final Answer (iii):

\frac{3\sqrt{2}-\sqrt{6}}{8}

(iv) $\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$ We know the values:

\sin 30^{\circ} = \frac{1}{2}

,

\tan 45^{\circ} = 1

,

\csc 60^{\circ} = \frac{2}{\sqrt{3}}

,

\sec 30^{\circ} = \frac{2}{\sqrt{3}}

,

\cos 60^{\circ} = \frac{1}{2}

,

\cot 45^{\circ} = 1

.

= \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1} = \frac{\frac{3}{2} - \frac{2}{\sqrt{3}}}{\frac{3}{2} + \frac{2}{\sqrt{3}}}

= \frac{\frac{3\sqrt{3}-4}{2\sqrt{3}}}{\frac{3\sqrt{3}+4}{2\sqrt{3}}} = \frac{3\sqrt{3}-4}{3\sqrt{3}+4}

To rationalize, multiply numerator and denominator by

(3\sqrt{3}-4)

:

= \frac{(3\sqrt{3}-4)(3\sqrt{3}-4)}{(3\sqrt{3}+4)(3\sqrt{3}-4)} = \frac{(3\sqrt{3}-4)^2}{(3\sqrt{3})^2 - 4^2}

= \frac{(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + 4^2}{27 - 16} = \frac{27 - 24\sqrt{3} + 16}{11} = \frac{43 - 24\sqrt{3}}{11}

Final Answer (iv):

\frac{43 - 24\sqrt{3}}{11}

(v) $\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$ We know the values:

\cos 60^{\circ} = \frac{1}{2}

,

\sec 30^{\circ} = \frac{2}{\sqrt{3}}

,

\tan 45^{\circ} = 1

. The denominator

\sin^2 30^{\circ} + \cos^2 30^{\circ} = 1

by the identity

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

.

= \frac{5(\frac{1}{2})^2 + 4(\frac{2}{\sqrt{3}})^2 - (1)^2}{1}

= 5\left(\frac{1}{4}\right) + 4\left(\frac{4}{3}\right) - 1

= \frac{5}{4} + \frac{16}{3} - 1 = \frac{5(3) + 16(4) - 1(12)}{12}

= \frac{15 + 64 - 12}{12} = \frac{67}{12}

Final Answer (v):

\frac{67}{12}

Q2EXERCISE 8.2

Choose the correct option and justify your choice :

(i)

$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=$

(A)

\sin 60^{\circ}

(B)

\cos 60^{\circ}

(C)

\tan 60^{\circ}

(D)

\sin 30^{\circ}

(ii)

$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$

(A)

\tan 90^{\circ}

(B)

1

(C)

\sin 45^{\circ}

(D)

0

(iii)

$\sin 2 \mathrm{~A}=2 \sin \mathrm{A}$ is true when $\mathrm{A}=$

(A)

0^{\circ}

(B)

30^{\circ}

(C)

45^{\circ}

(D)

60^{\circ}

(iv)

$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$

(A)

\cos 60^{\circ}

(B)

\sin 60^{\circ}

(C)

\tan 60^{\circ}

(D)

\sin 30^{\circ}

Solution

(i) $\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$ Solution: We know that

\tan 30^{\circ} = \frac{1}{\sqrt{3}}

.

\frac{2 \tan 30^{\circ}}{1+\tan^2 30^{\circ}} = \frac{2(\frac{1}{\sqrt{3}})}{1 + (\frac{1}{\sqrt{3}})^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{3+1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}

= \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3} \times \sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{3}}{2}

Now we check the options: (A)

\sin 60^{\circ} = \frac{\sqrt{3}}{2}

(B)

\cos 60^{\circ} = \frac{1}{2}

(C)

\tan 60^{\circ} = \sqrt{3}

(D)

\sin 30^{\circ} = \frac{1}{2}

The calculated value matches

\sin 60^{\circ}

. Answer: (A)

\sin 60^{\circ}

(ii) $\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$ Solution: We know that

\tan 45^{\circ} = 1

.

\frac{1 - \tan^2 45^{\circ}}{1 + \tan^2 45^{\circ}} = \frac{1 - (1)^2}{1 + (1)^2} = \frac{1-1}{1+1} = \frac{0}{2} = 0

Answer: (D) 0

**(iii)

\sin 2 \mathrm{~A}=2 \sin \mathrm{A}

is true when

\mathrm{A}=** **Solution:** We check the equation for each option. (A) For A =

0^{\circ}

: LHS:

\sin(2 \times 0^{\circ}) = \sin 0^{\circ} = 0

. RHS:

2 \sin 0^{\circ} = 2 \times 0 = 0

. LHS = RHS. So, it is true for A =

0^{\circ}

. (B) For A =

30^{\circ}

: LHS:

\sin(2 \times 30^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2}

. RHS:

2 \sin 30^{\circ} = 2 \times \frac{1}{2} = 1

. LHS

\neq

RHS. (C) For A =

45^{\circ}

: LHS:

\sin(2 \times 45^{\circ}) = \sin 90^{\circ} = 1

. RHS:

2 \sin 45^{\circ} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}

. LHS

\neq

RHS. (D) For A =

60^{\circ}

: LHS:

\sin(2 \times 60^{\circ}) = \sin 120^{\circ} = \frac{\sqrt{3}}{2}

. RHS:

2 \sin 60^{\circ} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}

. LHS

\neq

RHS. **Answer:** (A)

0^{\circ}$

(iv) $\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$ Solution: We know that

\tan 30^{\circ} = \frac{1}{\sqrt{3}}

.

\frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}} = \frac{2(\frac{1}{\sqrt{3}})}{1 - (\frac{1}{\sqrt{3}})^2} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}

= \frac{2}{\sqrt{3}} \times \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3}

Now we check the options: (A)

\cos 60^{\circ} = \frac{1}{2}

(B)

\sin 60^{\circ} = \frac{\sqrt{3}}{2}

(C)

\tan 60^{\circ} = \sqrt{3}

(D)

\sin 30^{\circ} = \frac{1}{2}

The calculated value matches

\tan 60^{\circ}

. Answer: (C)

\tan 60^{\circ}

Q3EXERCISE 8.2

If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}} ; 0^{\circ}B$ , find A and B.

Solution

Given:

\tan(A+B) = \sqrt{3}

\tan(A-B) = \frac{1}{\sqrt{3}}

0^{\circ} < A+B \leq 90^{\circ}

A > B

To Find: The values of A and B.

Solution: We know the standard values for the tangent function:

\tan 60^{\circ} = \sqrt{3}

\tan 30^{\circ} = \frac{1}{\sqrt{3}}

From the first given equation:

\tan(A+B) = \sqrt{3} = \tan 60^{\circ}

Since

A+B

is an acute angle, we can write:

A + B = 60^{\circ} \quad ...(1)

From the second given equation:

\tan(A-B) = \frac{1}{\sqrt{3}} = \tan 30^{\circ}

So, we can write:

A - B = 30^{\circ} \quad ...(2)

Now we have a system of two linear equations with two variables, A and B. We can solve them. Adding equation (1) and equation (2):

(A+B) + (A-B) = 60^{\circ} + 30^{\circ}

2A = 90^{\circ}

A = \frac{90^{\circ}}{2} = 45^{\circ}

Substitute the value of A into equation (1):

45^{\circ} + B = 60^{\circ}

B = 60^{\circ} - 45^{\circ}

B = 15^{\circ}

Let's check the conditions:

A+B = 45^{\circ} + 15^{\circ} = 60^{\circ}

, which is in the range

0^{\circ} < A+B \leq 90^{\circ}

. Also,

A=45^{\circ} > B=15^{\circ}

. The conditions are satisfied.

Final Answer: A =

45^{\circ}

and B =

15^{\circ}

.

Q4EXERCISE 8.2

State whether the following are true or false. Justify your answer.

(i)

$\sin (A+B)=\sin A+\sin B$ .

(ii)

The value of $\sin \theta$ increases as $\theta$ increases.

(iii)

The value of $\cos \theta$ increases as $\theta$ increases.

(iv)

$\sin \theta=\cos \theta$ for all values of $\theta$ .

(v)

$\cot \mathrm{A}$ is not defined for $\mathrm{A}=0^{\circ}$ .

Solution

(i) $\sin (A+B)=\sin A+\sin B$ . Answer: False. Reason: Let's take an example. Let A =

30^{\circ}

and B =

60^{\circ}

. LHS:

\sin(A+B) = \sin(30^{\circ} + 60^{\circ}) = \sin 90^{\circ} = 1

. RHS:

\sin A + \sin B = \sin 30^{\circ} + \sin 60^{\circ} = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}

. Since

1 \neq \frac{1+\sqrt{3}}{2}

, the statement is false.

(ii) The value of $\sin \theta$ increases as $\theta$ increases. Answer: True (for the range

0^{\circ} \leq \theta \leq 90^{\circ}

). Reason: Let's look at the values of

\sin \theta

as

\theta

increases from

0^{\circ}

to

90^{\circ}

:

\sin 0^{\circ} = 0

\sin 30^{\circ} = 0.5

\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707

\sin 60^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866

\sin 90^{\circ} = 1

As

\theta

increases from

0^{\circ}

to

90^{\circ}

, the value of

\sin \theta

increases from 0 to 1.

(iii) The value of $\cos \theta$ increases as $\theta$ increases. Answer: False (for the range

0^{\circ} \leq \theta \leq 90^{\circ}

). Reason: Let's look at the values of

\cos \theta

as

\theta

increases from

0^{\circ}

to

90^{\circ}

:

\cos 0^{\circ} = 1

\cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866

\cos 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707

\cos 60^{\circ} = 0.5

\cos 90^{\circ} = 0

As

\theta

increases from

0^{\circ}

to

90^{\circ}

, the value of

\cos \theta

decreases from 1 to 0.

(iv) $\sin \theta=\cos \theta$ for all values of $\theta$ . Answer: False. Reason: This equality is only true for

\theta = 45^{\circ}

in the range

0^{\circ} \leq \theta \leq 90^{\circ}

, where

\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}

. For other values, it is not true. For example,

\sin 30^{\circ} = \frac{1}{2}

while

\cos 30^{\circ} = \frac{\sqrt{3}}{2}

.

(v) $\cot \mathrm{A}$ is not defined for $\mathrm{A}=0^{\circ}$ . Answer: True. Reason: We know that

\cot A = \frac{\cos A}{\sin A}

. For A =

0^{\circ}

,

\cos 0^{\circ} = 1

and

\sin 0^{\circ} = 0

. So,

\cot 0^{\circ} = \frac{1}{0}

, which is not defined because division by zero is undefined.

Q1EXERCISE 8.3

Express the trigonometric ratios $\sin \mathrm{A}, \sec \mathrm{A}$ and $\tan \mathrm{A}$ in terms of $\cot \mathrm{A}$ .

Solution

To Express:

\sin A, \sec A, \tan A

in terms of

\cot A

.

Solution:

1. Expressing $\tan A$ in terms of $\cot A$ : This is a direct reciprocal relationship.

\tan A = \frac{1}{\cot A}

2. Expressing $\sin A$ in terms of $\cot A$ : We use the identity

\csc^2 A = 1 + \cot^2 A

. We know that

\sin A = \frac{1}{\csc A}

. From the identity,

\csc A = \pm \sqrt{1 + \cot^2 A}

. Since A is an acute angle,

\sin A

and

\csc A

are positive. So,

\csc A = \sqrt{1 + \cot^2 A}

. Therefore,

\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}

3. Expressing $\sec A$ in terms of $\cot A$ : We use the identity

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

. Substitute

\tan A = \frac{1}{\cot A}

into the identity:

\sec^2 A = 1 + \left(\frac{1}{\cot A}\right)^2 = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A}

Taking the square root, and since A is acute,

\sec A

is positive:

\sec A = \sqrt{\frac{\cot^2 A + 1}{\cot^2 A}} = \frac{\sqrt{\cot^2 A + 1}}{\cot A}

Final Answer:

\sin A = \frac{1}{\sqrt{1+\cot^2 A}}

\sec A = \frac{\sqrt{\cot^2 A+1}}{\cot A}

\tan A = \frac{1}{\cot A}

Q2EXERCISE 8.3

Write all the other trigonometric ratios of $\angle \mathrm{A}$ in terms of $\sec \mathrm{A}$ .

Solution

To Express:

\sin A, \cos A, \tan A, \cot A, \csc A

in terms of

\sec A

.

Solution:

1. Expressing $\cos A$ :

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

2. Expressing $\sin A$ : Using the identity

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

.

\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{1}{\sec A}\right)^2 = 1 - \frac{1}{\sec^2 A} = \frac{\sec^2 A - 1}{\sec^2 A}

Since A is an acute angle,

\sin A

is positive.

\sin A = \sqrt{\frac{\sec^2 A - 1}{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{\sec A}

3. Expressing $\tan A$ : Using the identity

\tan^2 A = \sec^2 A - 1

. Since A is an acute angle,

\tan A

is positive.

\tan A = \sqrt{\sec^2 A - 1}

4. Expressing $\csc A$ :

\csc A = \frac{1}{\sin A} = \frac{1}{\frac{\sqrt{\sec^2 A - 1}}{\sec A}} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}

5. Expressing $\cot A$ :

\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}

Final Answer:

\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}

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

\tan A = \sqrt{\sec^2 A - 1}

\csc A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}

\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}

Q3EXERCISE 8.3

Choose the correct option. Justify your choice.

(i)

$9 \sec ^{2} \mathrm{~A}-9 \tan ^{2} \mathrm{~A}=$

(A)

1

(B)

9

(C)

8

(D)

0

(ii)

$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=$

(A)

0

(B)

1

(C)

2

(D)

-1

(iii)

$(\sec \mathrm{A}+\tan \mathrm{A})(1-\sin \mathrm{A})=$

(A)

\sec \mathrm{A}

(B)

\sin \mathrm{A}

(C)

\operatorname{cosec} \mathrm{A}

(D)

\cos \mathrm{A}

(iv)

$\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}=$

(A)

\sec ^{2} \mathrm{~A}

(B)

-1

(C)

\cot ^{2} A

(D)

\tan ^{2} \mathrm{~A}

Solution

(i) $9 \sec ^{2} \mathrm{~A}-9 \tan ^{2} \mathrm{~A}$ Solution: Factor out the common term 9:

9(\sec^2 A - \tan^2 A)

Using the identity

\sec^2 A - \tan^2 A = 1

:

= 9(1) = 9

Answer: (B) 9

(ii) $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$ Solution: Convert all terms to

\sin \theta

and

\cos \theta

:

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

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

Let

(\cos \theta + \sin \theta) = x

. The numerator becomes

(x+1)(x-1) = x^2 - 1

.

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

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

Using the identity

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

:

= \frac{1 + 2\sin \theta \cos \theta - 1}{\cos \theta \sin \theta} = \frac{2\sin \theta \cos \theta}{\cos \theta \sin \theta} = 2

Answer: (C) 2

(iii) $(\sec \mathrm{A}+\tan \mathrm{A})(1-\sin \mathrm{A})$ Solution: Convert

\sec A

and

\tan A

to

\sin A

and

\cos A

:

= \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) (1 - \sin A)

= \left(\frac{1 + \sin A}{\cos A}\right) (1 - \sin A)

= \frac{(1 + \sin A)(1 - \sin A)}{\cos A} = \frac{1 - \sin^2 A}{\cos A}

Using the identity

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

:

= \frac{\cos^2 A}{\cos A} = \cos A

Answer: (D)

\cos \mathrm{A}

(iv) $\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}$ Solution: Using the identities

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

and

1 + \cot^2 A = \csc^2 A

:

= \frac{\sec^2 A}{\csc^2 A}

Convert to

\sin A

and

\cos A

:

= \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A

Answer: (D)

\tan ^{2} \mathrm{~A}

Q4EXERCISE 8.3

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i)

$(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$

(ii)

$\frac{\cos \mathrm{A}}{1+\sin \mathrm{A}}+\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}=2 \sec \mathrm{A}$

(iii)

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$

(iv)

$\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sin ^{2} \mathrm{~A}}{1-\cos \mathrm{A}}$

(v)

$\frac{\cos \mathrm{A}-\sin \mathrm{A}+1}{\cos \mathrm{A}+\sin \mathrm{A}-1}=\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}$ , using the identity $\operatorname{cosec}^{2} \mathrm{~A}=1+\cot ^{2} \mathrm{~A}$ .

(vi)

$\sqrt{\frac{1+\sin \mathrm{A}}{1-\sin \mathrm{A}}}=\sec \mathrm{A}+\tan \mathrm{A}$

(vii)

$\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta$

(viii)

$(\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^{2}+(\cos \mathrm{A}+\sec \mathrm{A})^{2}=7+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}$

(ix)

$(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$

(x)

$\left(\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}\right)=\left(\frac{1-\tan \mathrm{A}}{1-\cot \mathrm{A}}\right)^{2}=\tan ^{2} \mathrm{~A}$

Solution

(i) $(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$ To Prove: The identity is true. Proof: LHS =

(\csc \theta - \cot \theta)^2

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

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

Using

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

:

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

= \frac{1 - \cos \theta}{1 + \cos \theta} = \text{RHS}

Hence Proved.

(ii) $\frac{\cos \mathrm{A}}{1+\sin \mathrm{A}}+\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}=2 \sec \mathrm{A}$ To Prove: The identity is true. Proof: LHS =

\frac{\cos A}{1+\sin A} + \frac{1+\sin A}{\cos A}

Find a common denominator:

= \frac{\cos^2 A + (1+\sin A)^2}{(1+\sin A)\cos A}

= \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{(1+\sin A)\cos A}

Using

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

:

= \frac{( \cos^2 A + \sin^2 A) + 1 + 2\sin A}{(1+\sin A)\cos A} = \frac{1 + 1 + 2\sin A}{(1+\sin A)\cos A}

= \frac{2 + 2\sin A}{(1+\sin A)\cos A} = \frac{2(1+\sin A)}{(1+\sin A)\cos A}

= \frac{2}{\cos A} = 2 \sec A = \text{RHS}

Hence Proved.

(iii) $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$ To Prove: The identity is true. Proof: LHS =

\frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta}

Convert to

\sin \theta

and

\cos \theta

:

= \frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}

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

= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}

= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}

Using

a^3-b^3 = (a-b)(a^2+ab+b^2)

:

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

= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}

= \csc \theta \sec \theta + 1 = \text{RHS}

Hence Proved.

(iv) $\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sin ^{2} \mathrm{~A}}{1-\cos \mathrm{A}}$ To Prove: The identity is true. Proof: LHS =

\frac{1+\sec A}{\sec A} = \frac{1 + 1/\cos A}{1/\cos A} = \frac{(\cos A + 1)/\cos A}{1/\cos A} = \cos A + 1

RHS =

\frac{\sin^2 A}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A

Since LHS = RHS, Hence Proved.

(v) $\frac{\cos \mathrm{A}-\sin \mathrm{A}+1}{\cos \mathrm{A}+\sin \mathrm{A}-1}=\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}$ To Prove: The identity is true. Proof: LHS =

\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}

Divide numerator and denominator by

\sin A

:

= \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A}

= \frac{(\cot A + \csc A) - 1}{(\cot A - \csc A) + 1}

Using the identity

1 = \csc^2 A - \cot^2 A

in the numerator:

= \frac{(\cot A + \csc A) - (\csc^2 A - \cot^2 A)}{(\cot A - \csc A) + 1}

= \frac{(\csc A + \cot A) - (\csc A - \cot A)(\csc A + \cot A)}{\cot A - \csc A + 1}

Factor out

(\csc A + \cot A)

from the numerator:

= \frac{(\csc A + \cot A)[1 - (\csc A - \cot A)]}{\cot A - \csc A + 1}

= \frac{(\csc A + \cot A)(1 - \csc A + \cot A)}{1 - \csc A + \cot A}

= \csc A + \cot A = \text{RHS}

Hence Proved.

(vi) $\sqrt{\frac{1+\sin \mathrm{A}}{1-\sin \mathrm{A}}}=\sec \mathrm{A}+\tan \mathrm{A}$ To Prove: The identity is true. Proof: LHS =

\sqrt{\frac{1+\sin A}{1-\sin A}}

Multiply numerator and denominator inside the square root by

(1+\sin A)

:

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

= \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}} = \frac{1+\sin A}{\cos A}

= \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS}

Hence Proved.

(vii) $\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta$ To Prove: The identity is true. Proof: LHS =

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

Using

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

in the numerator:

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

= \frac{\sin \theta(2\cos^2 \theta - 1)}{\cos \theta(2\cos^2 \theta - 1)} = \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{RHS}

Hence Proved.

(viii) $(\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^{2}+(\cos \mathrm{A}+\sec \mathrm{A})^{2}=7+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}$ To Prove: The identity is true. Proof: LHS =

(\sin A + \csc A)^2 + (\cos A + \sec A)^2

= (\sin^2 A + \csc^2 A + 2\sin A \csc A) + (\cos^2 A + \sec^2 A + 2\cos A \sec A)

Since

\csc A = 1/\sin A

and

\sec A = 1/\cos A

:

= \sin^2 A + \csc^2 A + 2 + \cos^2 A + \sec^2 A + 2

= (\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 4

Using identities

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

,

\csc^2 A = 1 + \cot^2 A

,

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

:

= 1 + (1+\cot^2 A) + (1+\tan^2 A) + 4

= 1 + 1 + \cot^2 A + 1 + \tan^2 A + 4 = 7 + \tan^2 A + \cot^2 A = \text{RHS}

Hence Proved.

(ix) $(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$ To Prove: The identity is true. Proof: LHS =

(\csc A - \sin A)(\sec A - \cos A)

= \left(\frac{1}{\sin A} - \sin A\right) \left(\frac{1}{\cos A} - \cos A\right)

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

= \frac{\cos^2 A}{\sin A} \times \frac{\sin^2 A}{\cos A} = \sin A \cos A

RHS =

\frac{1}{\tan A + \cot A} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}

= \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \frac{1}{\frac{1}{\sin A \cos A}} = \sin A \cos A

Since LHS = RHS, Hence Proved.

(x) $\left(\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}\right)=\left(\frac{1-\tan \mathrm{A}}{1-\cot \mathrm{A}}\right)^{2}=\tan ^{2} \mathrm{~A}$ To Prove: The identity is true. Proof: First part:

\frac{1+\tan^2 A}{1+\cot^2 A} = \tan^2 A

LHS = \frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A = RHS

Second part:

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

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

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

Since both expressions equal

\tan^2 A

, the identity is true. Hence Proved.

Introduction to TrigonometryClass 10 Mathematics NCERT Solutions

In △ABC\triangle \mathrm{ABC}△ABC, right-angled at B,AB=24 cm,BC=7 cm\mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}, \mathrm{BC}=7 \mathrm{~cm}B,AB=24 cm,BC=7 cm. Determine :

sin⁡A,cos⁡A\sin \mathrm{A}, \cos \mathrm{A}sinA,cosA

sin⁡C,cos⁡C\sin \mathrm{C}, \cos \mathrm{C}sinC,cosC

Solution

In Fig. 8.13, find tan⁡P−cot⁡R\tan \mathrm{P}-\cot \mathrm{R}tanP−cotR.

Solution

If sin⁡A=34\sin \mathrm{A}=\frac{3}{4}sinA=43​, calculate cos⁡A\cos \mathrm{A}cosA and tan⁡A\tan \mathrm{A}tanA.

Solution

Given 15cot⁡A=815 \cot \mathrm{A}=815cotA=8, find sin⁡A\sin \mathrm{A}sinA and sec⁡A\sec \mathrm{A}secA.

Solution

Given sec⁡θ=1312\sec \theta=\frac{13}{12}secθ=1213​, calculate all other trigonometric ratios.

Solution

If ∠A\angle \mathrm{A}∠A and ∠B\angle \mathrm{B}∠B are acute angles such that cos⁡A=cos⁡B\cos \mathrm{A}=\cos \mathrm{B}cosA=cosB, then show that ∠A=∠B\angle \mathrm{A}=\angle \mathrm{B}∠A=∠B.

Solution

If cot⁡θ=78\cot \theta=\frac{7}{8}cotθ=87​, evaluate :

(1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}(1+cosθ)(1−cosθ)(1+sinθ)(1−sinθ)​,

cot⁡2θ\cot ^{2} \thetacot2θ

Solution

If 3cot⁡A=43 \cot \mathrm{A}=43cotA=4, check whether 1−tan⁡2 A1+tan⁡2 A=cos⁡2 A−sin⁡2 A\frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}1+tan2 A1−tan2 A​=cos2 A−sin2 A or not.

Solution

In triangle ABC , right-angled at B , if tan⁡A=13\tan \mathrm{A}=\frac{1}{\sqrt{3}}tanA=3​1​, find the value of:

sin⁡Acos⁡C+cos⁡Asin⁡C\sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C}sinAcosC+cosAsinC

cos⁡Acos⁡C−sin⁡Asin⁡C\cos \mathrm{A} \cos \mathrm{C}-\sin \mathrm{A} \sin \mathrm{C}cosAcosC−sinAsinC

Solution

In △PQR\triangle \mathrm{PQR}△PQR, right-angled at Q,PR+QR=25 cm\mathrm{Q}, \mathrm{PR}+\mathrm{QR}=25 \mathrm{~cm}Q,PR+QR=25 cm and PQ=5 cm\mathrm{PQ}=5 \mathrm{~cm}PQ=5 cm. Determine the values of sin⁡P,cos⁡P\sin \mathrm{P}, \cos \mathrm{P}sinP,cosP and tan P .

Solution

State whether the following are true or false. Justify your answer.

The value of tan⁡A\tan \mathrm{A}tanA is always less than 1 .

sec⁡A=125\sec \mathrm{A}=\frac{12}{5}secA=512​ for some value of angle A .

cos⁡A\cos \mathrm{A}cosA is the abbreviation used for the cosecant of angle A .

cot⁡A\cot \mathrm{A}cotA is the product of cot⁡\cotcot and A .

sin⁡θ=43\sin \theta=\frac{4}{3}sinθ=34​ for some angle θ\thetaθ.

Solution

Evaluate the following:

sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}sin60∘cos30∘+sin30∘cos60∘

2tan⁡245∘+cos⁡230∘−sin⁡260∘2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}2tan245∘+cos230∘−sin260∘

cos⁡45∘sec⁡30∘+cosec⁡30∘\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}sec30∘+cosec30∘cos45∘​

sin⁡30∘+tan⁡45∘−cosec⁡60∘sec⁡30∘+cos⁡60∘+cot⁡45∘\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}sec30∘+cos60∘+cot45∘sin30∘+tan45∘−cosec60∘​

5cos⁡260∘+4sec⁡230∘−tan⁡245∘sin⁡230∘+cos⁡230∘\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}sin230∘+cos230∘5cos260∘+4sec230∘−tan245∘​

Solution

Choose the correct option and justify your choice :

2tan⁡30∘1+tan⁡230∘=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=1+tan230∘2tan30∘​=

1−tan⁡245∘1+tan⁡245∘=\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=1+tan245∘1−tan245∘​=

sin⁡2 A=2sin⁡A\sin 2 \mathrm{~A}=2 \sin \mathrm{A}sin2 A=2sinA is true when A=\mathrm{A}=A=

2tan⁡30∘1−tan⁡230∘=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=1−tan230∘2tan30∘​=

Solution

If tan⁡(A+B)=3\tan (A+B)=\sqrt{3}tan(A+B)=3​ and tan⁡(A−B)=13;0∘B\tan (A-B)=\frac{1}{\sqrt{3}} ; 0^{\circ}Btan(A−B)=3​1​;0∘B, find A and B.

Solution

State whether the following are true or false. Justify your answer.

sin⁡(A+B)=sin⁡A+sin⁡B\sin (A+B)=\sin A+\sin Bsin(A+B)=sinA+sinB.

The value of sin⁡θ\sin \thetasinθ increases as θ\thetaθ increases.

The value of cos⁡θ\cos \thetacosθ increases as θ\thetaθ increases.

sin⁡θ=cos⁡θ\sin \theta=\cos \thetasinθ=cosθ for all values of θ\thetaθ.

cot⁡A\cot \mathrm{A}cotA is not defined for A=0∘\mathrm{A}=0^{\circ}A=0∘.

Solution

Express the trigonometric ratios sin⁡A,sec⁡A\sin \mathrm{A}, \sec \mathrm{A}sinA,secA and tan⁡A\tan \mathrm{A}tanA in terms of cot⁡A\cot \mathrm{A}cotA.

Solution

Write all the other trigonometric ratios of ∠A\angle \mathrm{A}∠A in terms of sec⁡A\sec \mathrm{A}secA.

Solution

Choose the correct option. Justify your choice.

9sec⁡2 A−9tan⁡2 A=9 \sec ^{2} \mathrm{~A}-9 \tan ^{2} \mathrm{~A}=9sec2 A−9tan2 A=

(1+tan⁡θ+sec⁡θ)(1+cot⁡θ−cosec⁡θ)=(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=(1+tanθ+secθ)(1+cotθ−cosecθ)=

(sec⁡A+tan⁡A)(1−sin⁡A)=(\sec \mathrm{A}+\tan \mathrm{A})(1-\sin \mathrm{A})=(secA+tanA)(1−sinA)=

1+tan⁡2 A1+cot⁡2 A=\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}=1+cot2 A1+tan2 A​=

Solution

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(cosec⁡θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}(cosecθ−cotθ)2=1+cosθ1−cosθ​

cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\frac{\cos \mathrm{A}}{1+\sin \mathrm{A}}+\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}=2 \sec \mathrm{A}1+sinAcosA​+cosA1+sinA​=2secA

tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θcosec⁡θ\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta1−cotθtanθ​+1−tanθcotθ​=1+secθcosecθ

1+sec⁡Asec⁡A=sin⁡2 A1−cos⁡A\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sin ^{2} \mathrm{~A}}{1-\cos \mathrm{A}}secA1+secA​=1−cosAsin2 A​

1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\sqrt{\frac{1+\sin \mathrm{A}}{1-\sin \mathrm{A}}}=\sec \mathrm{A}+\tan \mathrm{A}1−sinA1+sinA​​=secA+tanA

sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta2cos3θ−cosθsinθ−2sin3θ​=tanθ

(sin⁡A+cosec⁡A)2+(cos⁡A+sec⁡A)2=7+tan⁡2 A+cot⁡2 A(\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^{2}+(\cos \mathrm{A}+\sec \mathrm{A})^{2}=7+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}(sinA+cosecA)2+(cosA+secA)2=7+tan2 A+cot2 A

(cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}(cosecA−sinA)(secA−cosA)=tanA+cotA1​

(1+tan⁡2 A1+cot⁡2 A)=(1−tan⁡A1−cot⁡A)2=tan⁡2 A\left(\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}\right)=\left(\frac{1-\tan \mathrm{A}}{1-\cot \mathrm{A}}\right)^{2}=\tan ^{2} \mathrm{~A}(1+cot2 A1+tan2 A​)=(1−cotA1−tanA​)2=tan2 A

Solution

In $\triangle \mathrm{ABC}$ , right-angled at $\mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}, \mathrm{BC}=7 \mathrm{~cm}$ . Determine :

$\sin \mathrm{A}, \cos \mathrm{A}$

$\sin \mathrm{C}, \cos \mathrm{C}$

In Fig. 8.13, find $\tan \mathrm{P}-\cot \mathrm{R}$ .

If $\sin \mathrm{A}=\frac{3}{4}$ , calculate $\cos \mathrm{A}$ and $\tan \mathrm{A}$ .

Given $15 \cot \mathrm{A}=8$ , find $\sin \mathrm{A}$ and $\sec \mathrm{A}$ .

Given $\sec \theta=\frac{13}{12}$ , calculate all other trigonometric ratios.

If $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles such that $\cos \mathrm{A}=\cos \mathrm{B}$ , then show that $\angle \mathrm{A}=\angle \mathrm{B}$ .

If $\cot \theta=\frac{7}{8}$ , evaluate :

$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ ,

$\cot ^{2} \theta$

If $3 \cot \mathrm{A}=4$ , check whether $\frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}$ or not.

In triangle ABC , right-angled at B , if $\tan \mathrm{A}=\frac{1}{\sqrt{3}}$ , find the value of:

$\sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C}$

$\cos \mathrm{A} \cos \mathrm{C}-\sin \mathrm{A} \sin \mathrm{C}$

In $\triangle \mathrm{PQR}$ , right-angled at $\mathrm{Q}, \mathrm{PR}+\mathrm{QR}=25 \mathrm{~cm}$ and $\mathrm{PQ}=5 \mathrm{~cm}$ . Determine the values of $\sin \mathrm{P}, \cos \mathrm{P}$ and tan P .

The value of $\tan \mathrm{A}$ is always less than 1 .

$\sec \mathrm{A}=\frac{12}{5}$ for some value of angle A .

$\cos \mathrm{A}$ is the abbreviation used for the cosecant of angle A .

$\cot \mathrm{A}$ is the product of $\cot$ and A .

$\sin \theta=\frac{4}{3}$ for some angle $\theta$ .

$\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$

$2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$

$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$

$\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$

$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=$

$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$

$\sin 2 \mathrm{~A}=2 \sin \mathrm{A}$ is true when $\mathrm{A}=$

$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$

If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}} ; 0^{\circ}B$ , find A and B.

$\sin (A+B)=\sin A+\sin B$ .

The value of $\sin \theta$ increases as $\theta$ increases.

The value of $\cos \theta$ increases as $\theta$ increases.

$\sin \theta=\cos \theta$ for all values of $\theta$ .

$\cot \mathrm{A}$ is not defined for $\mathrm{A}=0^{\circ}$ .

Express the trigonometric ratios $\sin \mathrm{A}, \sec \mathrm{A}$ and $\tan \mathrm{A}$ in terms of $\cot \mathrm{A}$ .

Write all the other trigonometric ratios of $\angle \mathrm{A}$ in terms of $\sec \mathrm{A}$ .

$9 \sec ^{2} \mathrm{~A}-9 \tan ^{2} \mathrm{~A}=$

$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=$

$(\sec \mathrm{A}+\tan \mathrm{A})(1-\sin \mathrm{A})=$

$\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}=$

$(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$

$\frac{\cos \mathrm{A}}{1+\sin \mathrm{A}}+\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}=2 \sec \mathrm{A}$

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$

$\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sin ^{2} \mathrm{~A}}{1-\cos \mathrm{A}}$

$\sqrt{\frac{1+\sin \mathrm{A}}{1-\sin \mathrm{A}}}=\sec \mathrm{A}+\tan \mathrm{A}$

$\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta$

$(\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^{2}+(\cos \mathrm{A}+\sec \mathrm{A})^{2}=7+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}$

$(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$

$\left(\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}\right)=\left(\frac{1-\tan \mathrm{A}}{1-\cot \mathrm{A}}\right)^{2}=\tan ^{2} \mathrm{~A}$