# SSC > Trigonometric Ratios

Explore popular questions from Trigonometric Ratios for SSC. This collection covers Trigonometric Ratios previous year SSC questions hand picked by experienced teachers.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.

Correct2

Incorrect-0.5

If {tex} \sin \theta +cosec \theta =2,\ then\ \sin^{2} \theta + cosec^{2} \theta {/tex} is equal to

A

1

B

4

2

D

none of these

##### Explanation

Given that, {tex} \sin \theta +cosec \theta = 2 {/tex}

On squaring both sides, we get

{tex} \sin^{2} \theta +cosec^{2} \theta+2=4 {/tex}

=> {tex} \sin^{2} \theta + cosec^{2}\theta = 2 {/tex}

Q 2.

Correct2

Incorrect-0.5

The value of {tex} \frac{\cot^{2} \theta +1}{\cot^{2} \theta -1} {/tex} is equal to:

A

{tex} \sin 2 \theta {/tex}

B

{tex} \cos 2 \theta {/tex}

C

{tex} \cosh 2 \theta {/tex}

{tex} \sec 2 \theta {/tex}

##### Explanation

{tex} \frac{\cot^{2} \theta +1}{\cot^{2}\theta -1}=\frac{1+\tan^{2}\theta }{1-\tan^{2} \theta } {/tex}

={tex} \frac{1}{\cos^{2} \theta -\sin^{2} \theta } = \frac{1}{\cos 2 \theta } = \sec 2 \theta {/tex}

Q 3.

Correct2

Incorrect-0.5

If {tex} \tan A = 2 \tan B + \cot B,\ then\ 2 \tan \left(A-B\right) {/tex} is equal to:

A

{tex} \tan B {/tex}

B

{tex} 2 \tan B {/tex}

{tex} \cot B {/tex}

D

{tex} 2 \cot B {/tex}

##### Explanation

Given that

{tex} \tan A = 2 \tan B + \cot B {/tex} ...(i)

Now, {tex} 2 \tan \left(A-B\right)=2 \left(\frac{\tan A - \tan B}{1+ \tan A \tan B}\right) {/tex}

= {tex} 2 \frac{ \left(2 \tan B + \cot B - \tan B\right) }{1+ \left(2 \tan B + \cot B\right) \tan B } {/tex} From (i)

= {tex} 2 \frac{\tan B + \cot B}{2 \left(1+\tan^{2}B\right) } = \frac{\cot B \left( \tan^{2}B+1 \right) }{ \left(1+ \tan^{2}B\right) } = \cot B {/tex}

Q 4.

Correct2

Incorrect-0.5

If {tex} \tan A - \tan B = x\ and\ \cot B - \cot A = y,{/tex} then{tex} \cot \left(A-B\right) {/tex} is equal to:

A

{tex} \frac{1}{x}+y {/tex}

B

{tex} \frac{1}{xy} {/tex}

C

{tex} \frac{1}{x}-\frac{1}{y} {/tex}

{tex} \frac{1}{x}+\frac{1}{y} {/tex}

##### Explanation

Given that

{tex} \tan A - \tan B = x {/tex} ...(i)

and {tex} \cot B - \cot A = y {/tex} ...(ii)

Now, {tex} \cot \left(A-B\right) = \frac{1}{\tan \left(A-B\right) } {/tex}

= {tex} \frac{1+\tan A \tan B}{\tan A - \tan B} {/tex}

= {tex} \frac{1}{\tan A - \tan B} + \frac{\tan A \tan B}{\tan A - \tan B} {/tex}

= {tex} \frac{1}{x}+\frac{1}{y} {/tex} [from I and II]

Q 5.

Correct2

Incorrect-0.5

If {tex} \sin \theta = -\frac{4}{5}\ and\ \theta {/tex} and {tex} \theta {/tex} lies in the third quadrant, then {tex} \cos \frac{ \theta }{2} {/tex} is equal to:

A

{tex} \frac{1}{\sqrt{5}} {/tex}

{tex} -\frac{1}{\sqrt{5}} {/tex}

C

{tex} \frac{2}{\sqrt{5}} {/tex}

D

{tex} -\frac{2}{\sqrt{5}} {/tex}

##### Explanation

Given that,

{tex} \sin \theta = -\frac{4}{5} {/tex} and {tex} \theta {/tex} lies in the third quadrant

=> {tex} \cos \theta = -\sqrt{1-\frac{16}{25}} = -\frac{3}{5} {/tex}

Now, {tex} \cos \frac{ \theta }{2} = \pm \sqrt{\frac{1+\cos \theta }{2}} = \sqrt{\frac{1-\frac{3}{5}}{2}} = \pm \sqrt{\frac{1}{5}} {/tex}

But we take {tex} \cos \frac{ \theta }{2} = -\frac{1}{\sqrt{5}} {/tex}; Since, if {tex} \theta {/tex} lies in IIIrd quadrant, then {tex} \frac{ \theta }{2} {/tex} will be in IInd quadrant.

Hence, {tex} \cos \frac{ \theta }{2} = - \frac{1}{\sqrt{5}} {/tex}

Q 6.

Correct2

Incorrect-0.5

The value of {tex} \cos 1 ^{\circ} \cos 2 ^{\circ} \cos 3 ^{\circ} ....... \cos 100 ^{\circ} {/tex} is equal to:

A

1

B

-1

0

D

none of these

##### Explanation

{tex} \cos 1 ^{\circ} \cos 2 ^{\circ} \cos 3 ^{\circ} ....\cos 90 ^{\circ} .... \cos 100 ^{\circ} {/tex}

= {tex} \cos 1 ^{\circ} \cos 2 ^{\circ} \cos 3 ^{\circ} ....0 .... \cos 100 ^{\circ} = 0 {/tex}

{tex} [\because \cos 90 ^{\circ} = 0 ] {/tex}

Q 7.

Correct2

Incorrect-0.5

The value of {tex} \sin 12 ^{\circ} \sin 48 ^{\circ} \sin 54 ^{\circ} {/tex} is equal to:

A

{tex} \frac{1}{16} {/tex}

B

{tex} \frac{1}{32} {/tex}

{tex} \frac{1}{8} {/tex}

D

{tex} \frac{1}{4} {/tex}

##### Explanation

Now, {tex} \sin 12 ^{\circ} \sin 48 ^{\circ} \sin 54 ^{\circ} {/tex}

{tex} = \frac{1}{2} \left(\cos 36 ^{\circ} -\cos 60 ^{\circ} \right) \cos 36 ^{\circ} {/tex}

= {tex} \frac{1}{2}\left[ \frac{\sqrt{5}+1}{4}-\frac{1}{2} \right]\left[ \frac{\sqrt{5}+1}{4} \right]=\frac{5-1}{32}=\frac{4}{32}=\frac{1}{8} {/tex}

Q 8.

Correct2

Incorrect-0.5

The value of {tex} 2 cos\ x - \cos 3x - \cos 5x {/tex} is equal to

{tex} 16 \cos^{3} x \sin^{2} x {/tex}

B

{tex} 16 \sin^{3} x \cos^{2} x {/tex}

C

{tex} 4 \cos^{3} x \sin^{2} x {/tex}

D

{tex} 4 \sin^{3} x \cos^{2} x {/tex}

##### Explanation

{tex} 2 \cos x - \cos 3x - \cos 5x = 2 \cos x - 2 \cos x \cos 4x {/tex}

= {tex} 2 \cos x \left(1- \cos 4x\right) = 2 \cos x 2 \sin^{2} 2x {/tex}

= {tex} 4 \cos x \left(2 \sin x \cos\ x\right)^{2} = 16 \sin^{2}x \cos^{3}x {/tex}

Q 9.

Correct2

Incorrect-0.5

If {tex} \cos \theta = \frac{1}{2} \left(x+\frac{1}{x}\right),\ then\ \frac{1}{2} \left(x^{2} + \frac{1}{x^{2}}\right) {/tex} is equal to:

A

{tex} \sin 2 \theta {/tex}

B

{tex} \sec 2 \theta {/tex}

C

{tex} \tan 2 \theta {/tex}

{tex} \cos 2 \theta {/tex}

##### Explanation

Given that {tex} \cos \theta = \frac{1}{2} \left(x+\frac{1}{x}\right) => x+\frac{1}{x}=2 \cos \theta {/tex}

We know that {tex} x^{2}+\frac{1}{x^{2}} = \left(x+\frac{1}{x}\right)^{2}-2 {/tex}

= {tex} \left(2 \cos \theta \right)^{2}-2=4 \cos^{2} \theta -2 {/tex}

= {tex} 2 \cos 2 \theta {/tex} .. From (i)

Therefore, {tex} \frac{1}{2} \left(x^{2}+\frac{1}{x^{2}}\right)=\frac{1}{2} \times 2 \cos \theta = \cos 2 \theta {/tex}

Q 10.

Correct2

Incorrect-0.5

The value of x for the maximum value {tex} \sqrt{3} \cos x + \sin x {/tex} is:

{tex} 30 ^{\circ} {/tex}

B

{tex} 45 ^{\circ} {/tex}

C

{tex} 60 ^{\circ} {/tex}

D

{tex} 90 ^{\circ} {/tex}

##### Explanation

Let {tex} f \left(x\right)=\sqrt{3} \cos x + \sin x {/tex}

=> {tex} f \left(x\right)=2 \left(\frac{\sqrt{3}}{2}\cos x + \frac{1}{2} \sin x\right)=2 \sin \left(x+\frac{ \pi }{3}\right) {/tex}

Since, {tex} -1 \le \sin \left(x+\frac{ \pi }{3}\right) \le 1 {/tex}

Hence, {tex} f \left(x\right)\ is\ maximum\ if\ x+\frac{ \pi }{3}=\frac{ \pi }{2} {/tex}

=> {tex} x = \frac{ \pi }{6} = 30 ^{\circ} {/tex}

Q 11.

Correct2

Incorrect-0.5

The equation {tex} \left(a+b\right)^{2}=4ab \sin^{2} \theta {/tex} is possible only when

{tex} a = b {/tex}

B

{tex} 2a = b {/tex}

C

{tex} a = 2b {/tex}

D

none of these

##### Explanation

We have {tex} \left(a+b\right)^{2} = 4ab \sin^{2} \theta {/tex}

=> {tex} \sin^{2} \theta = \frac{ \left(a+b\right)^{2} }{4ab} {/tex}

Since, {tex} \sin^{2} \theta \le 1 {/tex}

=> {tex} \frac{ \left(a+b\right)^{2} }{4ab} \le 1 {/tex}

=> {tex} \left(a+b\right)^{2}-4ab \le 1 {/tex}

=> {tex} \left(a-b\right)^{2} \le 0 => a=b {/tex}

Q 12.

Correct2

Incorrect-0.5

The value of the expression {tex} 1 - \frac{\sin^{2} y}{1+\cos y}+\frac{1 + \cos y}{\sin y}-\frac{\sin y}{1- \cos y} {/tex} is equal to;

A

0

B

1

C

{tex} \sin y {/tex}

{tex} \cos y {/tex}

##### Explanation

The given expression can be written as

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

= {tex} \frac{\cos y \left(1+\cos y\right) }{1+\cos y}+0= \cos y {/tex}

Q 13.

Correct2

Incorrect-0.5

The circular wire of diameter 10 cm is cut and placed along the circumference of a circle of diameter 1m. The angle subtended by the write at the centre of the circle is equal to:

A

B

D

##### Explanation

Given that, diameter of circular wire = 10 cm,

Length of wire = {tex} 10 \pi {/tex}

Hence, required angle {tex} = \frac{length\ of\ arc }{radius\ of\ big\ circle } {/tex}

{tex} = \frac{10 \pi }{50}=\frac{ \pi }{5} rad {/tex}

Q 14.

Correct2

Incorrect-0.5

The greatest and least value of {tex} \sin x \cos\ x {/tex} are:

A

1, -1

{tex} \frac{1}{2},\ -\frac{1}{2} {/tex}

C

{tex} \frac{1}{4},\ -\frac{1}{4} {/tex}

D

2, -2

##### Explanation

Let {tex} f \left(x\right)= \sin x \cos x = \frac{1}{2}\sin 2x {/tex}

We know {tex} -1 \le \sin 2x \le 1 {/tex}

=> {tex} -\frac{1}{2} \le \frac{1}{2}\sin 2x \le \frac{1}{2} {/tex}

Thus, the greatest and least value of {tex} f \left(x\right) {/tex} are

{tex} \frac{1}{2}\ and\ -\frac{1}{2}{/tex}. respectively.

Q 15.

Correct2

Incorrect-0.5

If {tex} A = \sin^{2} \theta + \cos^{4} \theta {/tex}, then for all real values of {tex} \theta {/tex}

A

{tex} 1 \le A \le 2 {/tex}

{tex} \frac{3}{4} \le A \le 1 {/tex}

C

{tex} \frac{13}{16} \le A \le 1 {/tex}

D

{tex} \frac{3}{4} \le A \le \frac{13}{16} {/tex}

##### Explanation

We have, {tex} A = \sin^{2} \theta + \cos^{4} \theta {/tex}

= {tex} \sin^{2} \theta + \cos^{2} \theta \cos^{2} \theta \le \sin^{2} \theta + \cos^{2} \theta {/tex} (Since, {tex} \cos^{2} \theta \le 1 {/tex})

=> {tex} \sin^{2} \theta + \cos^{4} \theta \le 1 => A \le 1 {/tex}

Again, {tex} \sin^{2} \theta + \cos^{4} \theta =1 - \cos^{2} \theta + \cos^{4} \theta {/tex}

= {tex} \cos^{4} \theta - \cos^{2} \theta + 1 {/tex}

= {tex} \left(\cos^{2} \theta - \frac{1}{2}\right)^{2}+\frac{3}{4}\ge \frac{3}{4} {/tex}

Hence {tex} \frac{3}{4} \le A \le 1 {/tex}

Q 16.

Correct2

Incorrect-0.5

{tex} \tan \frac{2 \pi }{5} - \tan \frac{ \pi }{15} - \sqrt{3} \tan \frac{2 \pi }{5} \tan \frac{ \pi }{15} {/tex} is equal to:

A

{tex} -\sqrt{3} {/tex}

B

{tex} \frac{1}{\sqrt{3}} {/tex}

C

{tex} 1 {/tex}

{tex} \sqrt{3} {/tex}

##### Explanation

Now, {tex} \tan \frac{ \pi }{3}=\tan \left(\frac{6 \pi }{15}-\frac{ \pi }{15}\right)= \frac{\tan\frac{6 \pi }{15}-\tan\frac{ \pi }{15}}{1+\tan\frac{6 \pi }{15}\tan\frac{ \pi }{15}} {/tex}

=> {tex} \tan\frac{6 \pi }{15}-\tan\frac{ \pi }{15}=\sqrt{3}+\sqrt{3}\tan\frac{6 \pi }{15}\tan\frac{ \pi }{15} {/tex}

=> {tex} \tan\frac{6 \pi }{15}-\tan\frac{ \pi }{15}-\sqrt{3}\tan\frac{6 \pi }{15}\tan\frac{ \pi }{15}=\sqrt{3} {/tex}

=> {tex} \tan\frac{2 \pi }{5}-\tan\frac{ \pi }{15}-\sqrt{3}\tan\frac{2 \pi }{5}\tan\frac{ \pi }{15}=\sqrt{3} {/tex}

Q 17.

Correct2

Incorrect-0.5

The value of {tex} \tan \alpha + 2 \tan 2 \alpha + 4 \tan + 8 \cot 8 \alpha {/tex} is equal to:

A

{tex} \tan \alpha {/tex}

B

{tex} 2 \tan \alpha {/tex}

{tex} \cot \alpha {/tex}

D

{tex} \cot 2 \alpha {/tex}

##### Explanation

We have {tex} \tan \alpha + 2 \tan 2 \alpha + 4 \tan + 8 \cot 8 \alpha {/tex}

= {tex} \tan \alpha + 2 \tan 2 \alpha + 4 \left(\frac{\sin 4 \alpha }{\cos 4 \alpha }+2\frac{\cos 8 \alpha }{\sin 8 \alpha }\right) {/tex}

= {tex} \tan \alpha + 2 \tan \alpha + 4 \left[\frac{\cos 4 \cos 8 \alpha + \sin 4 \alpha \sin 8 \alpha + \cos 4 \alpha \cos 8 \alpha}{\sin 8 \alpha \cos 4 \alpha } \right] {/tex}

= {tex} \tan \alpha - 2 \tan 2 \alpha + 4 \left[ \frac{\cos 4 \alpha + \cos 4 \alpha \cos 8 \alpha }{\sin 8 \alpha \cos 4 \alpha } \right] {/tex}

= {tex} \tan \alpha + 2 \tan 2 \alpha + 4 \left[ \frac{\cos 4 \alpha \left(1+\cos 8 \alpha \right)}{\cos 4 \alpha \sin 8 \alpha } \right] {/tex}

= {tex} \tan \alpha + 2 \tan 2 \alpha + 4 \left[ \frac{2 \cos^{2} 4 \alpha }{2 \sin 4 \alpha \cos 4 \alpha } \right]{/tex}

= {tex} \tan \alpha + 2 \left(\tan \alpha + 2 \cot 4 \alpha \right) {/tex}

= {tex} \tan \alpha + 2 \left[ \frac{\sin 2 \alpha }{\cos 2 \alpha }+2\frac{\cos 4 \alpha }{\sin 4 \alpha } \right] {/tex}

= {tex} \tan \alpha +2\left[ \frac{\sin 2 \alpha \sin 4 + \cos 4 \alpha \cos 2 \alpha + \cos 4 \alpha \cos 2 \alpha }{\sin 4 \alpha \cos 2 \alpha } \right] {/tex}

= {tex} \tan \alpha +2 \left[ \frac{\cos 2 \alpha + \cos 2 \alpha \cos 4 \alpha}{\sin 4 \alpha \cos 2 \alpha } \right] {/tex}

= {tex} \tan \alpha +2 \left[ \frac{\cos 2 \alpha \left(1+\cos 4 \alpha \right) }{\sin 4 \alpha \cos 2 \alpha } \right] {/tex}

= {tex} \tan \alpha +2 \cot 2 \alpha = \frac{\sin \alpha }{\cos \alpha }+\frac{2 \cos 2 \alpha }{\sin 2 \alpha } = \frac{\cos \alpha + cos \alpha \cos 2 \alpha }{\sin 2 \alpha \cos \alpha } {/tex}

= {tex} \frac{1+\cos 2 \alpha }{\sin 2 \alpha } = \frac{2\cos^{2} \alpha }{2 \sin \alpha \cos \alpha } = \cot \alpha {/tex}

Q 18.

Correct2

Incorrect-0.5

If {tex} \alpha {/tex}, {tex} \beta {/tex} and {tex} \gamma {/tex} are acute angled such that {tex} sin \ \alpha=\frac{\sqrt{3}}{2} \ cos \ \beta=\frac{\sqrt{3}}{2} {/tex} and tan r=1 then what is {tex} \alpha + \beta + \gamma {/tex} equal to?

A

105 degree

B

120 degree

135 degree

D

150 degree

##### Explanation

{tex} sin \ \alpha =\frac{\sqrt{3}}{2} {/tex}

{tex} \alpha =60 ^{\circ} {/tex}

{tex} cos \ \beta =\frac{\sqrt{3}}{2} {/tex}

{tex} \beta =30 ^{\circ} {/tex}

{tex} tan \gamma=1 {/tex}

{tex} \gamma=45 ^{\circ} {/tex}

so, {tex} \alpha + \beta + \gamma=60 ^{\circ} +30 ^{\circ} +45 ^{\circ} =135 ^{\circ} {/tex}

Q 19.

Correct2

Incorrect-0.5

If {tex} sin \ \theta =\frac{a^{2}-1}{a^{2}+1} {/tex} then the value of {tex} sec \ \theta + tan \ \theta {/tex} will be

A

{tex} \frac{a}{\sqrt{2}} {/tex}

B

{tex} \frac{a}{a^{2}+1} {/tex}

C

{tex} \sqrt{2}a {/tex}

a

##### Explanation

{tex} sin \theta =\frac{a^{2}-1}{a^{2}+1} {/tex}

In, {tex} \triangle ABC {/tex}

{tex} BC=\sqrt{AC^{2}-AB^{2}} {/tex}

= {tex} \sqrt{(a^{2}+1)^{2}-(a^{2}-1)^{2}} {/tex}

= {tex} \sqrt{a^{2}+1+2a^{2}-a^{2}-1+2a^{2}} {/tex}

= {tex} \sqrt{4a^{2}}=2a {/tex}

{tex} sec \theta +tan \theta =\frac{a^{2}+1}{2a}+\frac{a^{2}-1}{2a} {/tex}

={tex} \frac{2a^{2}}{2a}=a {/tex}

Q 20.

Correct2

Incorrect-0.5

If {tex} \alpha + \beta {/tex}=90 degree and {tex} \alpha : \beta {/tex}=2:1 then the value of {tex} sin \alpha : sin \beta {/tex} is

{tex} \sqrt{3}:1 {/tex}

B

2 : 1

C

1 : 1

D

{tex} \sqrt{2}:1 {/tex}

##### Explanation

Given,

{tex} \alpha + \beta =90 ^{\circ} {/tex} and {tex} \alpha : \beta =2:1 {/tex}

{tex} \alpha =\frac{2}{2+1} \times 90 ^{\circ} {/tex}

= {tex} \frac{2}{3} \times 90 ^{\circ} =60 ^{\circ} {/tex}

Similarly, {tex} \beta =\frac{1}{3} \times 90 ^{\circ} =30 ^{\circ} {/tex}

{tex} sin \alpha :sin \beta =sin60 ^{\circ} :sin30 ^{\circ} {/tex}

= {tex} \frac{\sqrt{3}}{2}:\frac{1}{2}=\sqrt{3}:1 {/tex}

Q 21.

Correct2

Incorrect-0.5

The value of {tex} \left(\frac{1}{(1+tan^{2} \theta )}+\frac{1}{(1+cot^{2} \theta )}\right) {/tex} is

1

B

2

C

{tex} \frac{1}{2} {/tex}

D

{tex} \frac{1}{4} {/tex}

##### Explanation

{tex} \frac{1}{1+tan^{2} \theta }+\frac{1}{1+cot^{2} \theta } {/tex}

{tex} \frac{1}{1+\frac{sin^{2} \theta }{cos^{2} \theta }}+\frac{1}{1+\frac{cos^{2} \theta }{sin^{2} \theta }} {/tex}

= {tex} \frac{cos^{2} \theta }{cos^{2} \theta +sin^{2} \theta }+\frac{sin^{2} \theta }{sin^{2} \theta +cos^{2} \theta } {/tex}

{tex} cos^{2} \theta +sin^{2} \theta =1 {/tex}

Q 22.

Correct2

Incorrect-0.5

The minimum value of {tex} sin^{2} \theta+cos^{4} \theta {/tex} is

A

{tex} \frac{1}{\sqrt{2}} {/tex}

B

{tex} \frac{3}{5} {/tex}

{tex} \frac{3}{4} {/tex}

D

{tex} \frac{2}{3} {/tex}

##### Explanation

Minimum value of

{tex} sin^{2} \theta +cos^{4} \theta {/tex} will be at {tex} \theta {/tex} = 45 degree

Required minimum value

{tex} sin^{2} 45+cos^{4}45=\frac{1}{2}+\frac{1}{4}=\frac{3}{4} {/tex}

Q 23.

Correct2

Incorrect-0.5

If {tex} sec \theta +tan \theta =2 {/tex}, what is the value of {tex} sec \theta {/tex} ?

A

{tex} \frac{3}{2} {/tex}

B

{tex} \sqrt{2} {/tex}

C

{tex} \frac{5}{2} {/tex}

{tex} \frac{5}{4} {/tex}

##### Explanation

By trigonometric identity,

{tex} sec^{2} \theta -tan^{2} \theta =1 {/tex}

=> {tex} (sec \theta +tan \theta )(sec \theta -tan \theta )=1 {/tex}

=> {tex} sec \theta -tan \theta =\frac{1}{2} {/tex} ....(i)

and given, {tex} sec \theta +tan \theta =2 {/tex} .....(ii)

On adding Eqs. (i) and (ii), we get

{tex} 2sec \theta=\frac{1}{2}+2 {/tex}

{tex} sec \theta =\frac{5}{4} {/tex}

Q 24.

Correct2

Incorrect-0.5

What is {tex} sin 25^{\circ} sin 35^{\circ} sec 65^{\circ} sec 55^{\circ}{/tex} equal to?

A

-1

B

0

C

{tex} \frac{1}{2} {/tex}

1

##### Explanation

sin 25 {tex} ^{\circ}{/tex} sin 35 {tex} ^{\circ}{/tex} sec 65 {tex} ^{\circ}{/tex} sec 55 {tex} ^{\circ}{/tex}

=> {tex} sin \ 25 ^{\circ} \ sin \ 35 ^{\circ} .\frac{1}{cos \ 65 ^{\circ} }.\frac{1}{cos \ 55 ^{\circ} } {/tex}

=> {tex} sin \ 25 ^{\circ} . \ sin \ 35 ^{\circ} \times \frac{1}{cos(90 ^{\circ} -25 ^{\circ} )} \times \frac{1}{cos(90 ^{\circ} -35 ^{\circ} )} {/tex}

=> {tex} sin \ 25 ^{\circ} \times sin \ 35 ^{\circ} \times \frac{1}{sin25 ^{\circ} } \times \frac{1}{sin35 ^{\circ} } {/tex}

=> 1

Q 25.

Correct2

Incorrect-0.5

If p = a sin x + b cos x and q = a cos x - b sin x, then what is the of {tex} p ^{2}+q^{2} {/tex} ?

A

a + b

B

ab

{tex} a ^{2}+b^{2} {/tex}

D

{tex} a ^{2}-b^{2} {/tex}

##### Explanation

Given, p = asin x + bcos x ...(i)

and q = acos x - bsin x ....(ii)

On squaring both the equations

{tex} p^{2}=a^{2}sin^{2}x+b^{2}cos^{2}x {/tex}+2ab sinx cosx

and {tex} q^{2}=a^{2}cos^{2}x+b^{2}sin^{2}x {/tex}-2ab sinx cosx

{tex} p^{2}+q^{2}=a^{2}sin^{2}x+b^{2}cos^{2}x {/tex}

{tex} +2ab \ sinx \ cosx

+a^{2} \ cos^{2}x+b^{2}sin^{2}x-2ab \ sinx \ cosx {/tex}

{tex} p^{2}+q^{2}=a^{2}(sin^{2}x+cos^{2}x){/tex}

{tex}+b^{2}(cos^{2}x+sin^{2}x) {/tex}

= {tex} a^{2}+b^{2} {/tex}