SSC > Trigonometric Ratios

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}

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}

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}

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]

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}

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}

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}

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}

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}

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}

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}

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}

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}

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.

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}

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}

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}

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}

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}

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}

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}

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}

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}

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

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}