Mathematics

Trigonometric Identities

82 Questions

Trigonometric identities focus on solving equations using tangent, sine, and cosine properties. Questions involve half angle formulas and slope calculations. This topic is a staple in the quantitative aptitude section of major competitive examinations.

Tangent propertiesAngle formulasTrigonometric equationsHalf angle identitiesSlope calculations

Trigonometric Identities Questions

Multiple choice maths compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric ratios of acute angles trigonometric identities trigonometrical ratio and identities

If $tan x + cot x = 2$, then $sin^{2n}x+cos^{2n}x=$

  1. $\dfrac{1}{2}$
  2. $2^n$
  3. $\dfrac{1}{2^n}$
  4. $\dfrac{1}{2^{n-1}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $\tan x+\cot x=2$

$\implies \tan x+\dfrac{1}{\tan x}=2$
$\implies \tan^2 x-2\tan x+1=0$
$\implies (\tan x-1)^2=0$
$\implies \tan x=1\implies x=\dfrac{\pi}{4}$
$\sin^{2 n} x+\cos^{2 n} x=\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2 n}+\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2 n}=\dfrac{1}{2^n}+\dfrac{1}{2^n}=\dfrac{2}{2^n}=\dfrac{1}{2^{n-1}}$

Multiple choice maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

If $\tan x =\dfrac{3}{4} , \pi < x < \dfrac{3\pi}{2} $ find value of $\sin\dfrac{x}{2} , \cos\dfrac{x}{2} , \tan \dfrac{x}{2}$

  1. cos x/2= 3/2, sin x/2 =-3/2, tan x/2= -1/√10

  2. cos x/2= 3/√10, sin x/2 =-3/2, tan x/2= -1/√10

  3. cos x/2= -1/√10, sin x/2 = 3/√10, tan x/2= -3

  4. cos x/2= 3/√10, sin x/2 =-2/3, tan x/2= -1/√10
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given 

$ \pi < x < \dfrac {3\pi} 2$ 

$ \implies \dfrac \pi 2 < \dfrac x2 < \dfrac {3\pi }4 $ 

$ \implies \cos \dfrac x2 <0 , \sin \dfrac x2 >0 $ 

$ \tan x =\dfrac 34 $ 

$ \dfrac {2 \tan \dfrac x2 }{1-\tan ^2 \dfrac x2 }=\dfrac 34$

$ 8 \tan \dfrac x2 =3-3\tan ^2 \dfrac x2 $

$ 3\tan ^2 \dfrac x2 +8\tan \dfrac x2 -3=0 $ 

$ 3\tan ^2 \dfrac x2 +9\tan \dfrac x2 - \left(\tan \dfrac x2 +3 \right)=0 $ 

$ \tan \dfrac x2 =-3,\dfrac 13 $ 

As $ tan \dfrac x2 <0 \implies \tan \dfrac x2 =-3$ 

$ \cos \dfrac x2 =\dfrac {-1}{\sqrt {(-1)^2+3^2}}=\dfrac {-1}{\sqrt {10}}$ 

$ \sin \dfrac x2 =\dfrac {3}{\sqrt {(-1)^2+3^2}}=\dfrac {3}{\sqrt {10}}$ 

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $\tan 4x+\tan 5x-\tan 9x=k\tan 4x\tan 5x\tan 9x$ then $k=$

  1. $1$
  2. $-1$
  3. $ \pm 1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l}\tan 9x = \tan (4x + 5x) = \dfrac{{\tan 4x + \tan 5x}}{{1 - \tan 4x\tan 5x}}\ \Rightarrow \tan 9x - \tan 4x\tan 5x\tan 9x = \tan 4x + \tan 5x\ \Rightarrow \tan 4x + \tan 5x - \tan 9x =  - \tan 4x\tan 5x\tan 9x\\therefore k =  - 1\end{array}$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

State true or false $\tan(\dfrac{\pi}{4} + \theta) - \tan(\dfrac{\pi}{4} -\theta) = 2\tan\theta$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using $\tan (A+B)=\cfrac {\tan A+\tan B}{1-\tan A \tan B}$

$\Rightarrow \tan (A-B)=\cfrac {\tan A-\tan B}{1+\tan A+\tan B}$
$\tan \left( \cfrac {\pi}{4}+\theta\right)-\tan \left(\cfrac {\pi}{4}-\theta\right)$
$\Rightarrow A=\cfrac {\pi}{4}, B=\theta$
$\Rightarrow \cfrac {\tan \cfrac {\pi}{4}+\tan\theta}{1-\tan \cfrac {\pi}{4}\tan \theta}-\cfrac {\tan \cfrac {\pi}{4}-\tan \theta}{1+\tan \cfrac {\pi}{4}\tan \theta}$
$\because \tan \cfrac {\pi}{4}=1$
$\Rightarrow \cfrac {1+\tan\theta}{1-\tan \theta}-\cfrac {1-\tan \theta}{1+\tan \theta}$
$\Rightarrow \cfrac {(1+\tan\theta)^2-(1-\tan\theta)^2}{(1-\tan\theta)(1+\tan\theta)}$
$\therefore a^2-b^2=(a+b)(a-b)$
$\Rightarrow \cfrac {1+\tan^2\theta+2\tan\theta-1-\tan^2\theta+2\tan\theta}{1-\tan^2\theta}$
$\Rightarrow \cfrac {4\tan\theta}{1-\tan^2\theta}$
$\because \tan2\theta=\cfrac {2\tan\theta}{1-\tan^2\theta}$
$=2\tan2\theta$.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

$tan  5x-tan  3x-tan  2x=$

  1. $\tan 5x \tan 3x \tan 2x$
  2. $\sin 5x \sin 3x \sin 2x$
  3. $\cos 5x \cos 3x \cos 2x$
  4. $\sec 5x \sec 3x \sec 2x$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We've,

$\tan (3x+2x)=\tan 5x$

or, $\dfrac{\tan 3x+\tan 2x}{1-\tan 3x.\tan 2x}=\tan 5x$

or, $\tan 3x+\tan 2x=\tan 5x-\tan 2x.\tan 3x.\tan 5x$

or, $\tan 5x-\tan 3x-\tan 2x=\tan 2x.\tan 3x.\tan 5x$.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $\dfrac{\pi}{4}<A<\dfrac{\pi}{2}$ then $\tan^{-1}\left(\dfrac{1}{2}\tan 2A\right)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3}A)$=

  1. $0$
  2. $\pi$
  3. $\pi/2$
  4. $\pi/4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using trigonometric identities for inverse functions and the given range, the sum simplifies to 0. Specifically, the terms cancel out based on the properties of inverse tangents of cotangent functions.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $A+B+C=\pi $ and cosA=cosB cosC, then tanB tanC is equal to 

  1. $\frac { 1 }{ 2 } $
  2. $2$
  3. $1$
  4. $-\frac { 1 }{ 2 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $A+B+C=\pi\implies A=\pi-(B+C)$

And also given $\cos A=\cos B\cos C$
$\implies \cos (\pi-(B+C))=\cos B\cos C$
$\implies -\cos B\cos C+\sin B\sin C=\cos B\cos C$
$\implies \sin B\sin C=2\cos B\cos C$
$\implies \tan B\tan C=2$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

$\tan \alpha  + 2\tan 2\alpha  + 4\tan 4\alpha  + 8\tan 8\alpha  + 16\tan 16\alpha  + 32\cot 32\alpha $ is equal

  1. $\cot \alpha $
  2. $\tan \alpha $
  3. $\cos \alpha $
  4. $sin \alpha $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a telescoping series of trigonometric functions. Using the identity tan(x) = cot(x) - 2cot(2x), the sum collapses to cot(alpha).

Multiple choice trigonometric equations trigonometric functions trigonometry maths

General solution of $\dfrac{1-{tan}^{2}x}{{sec}^{2}x}=\dfrac{1}{2}$ is

  1. $n\pi+\dfrac{\pi}{6},n\in Z$
  2. $n\pi-\dfrac{\pi}{6},n\in Z$
  3. $n\pi\pm\dfrac{\pi}{6},n\in Z$
  4. $2n\pi\pm\dfrac{\pi}{6},n\in Z$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $\dfrac{1-\tan^2 x}{\text{sec}^2 x}=\dfrac{1}{2}$


$\implies \dfrac{1-\tan^2 x}{1+\tan^2 x}=\dfrac{1}{2}$


$\implies 2-2\tan^2 x=1+\tan^2 x$

$\implies \tan^2 x=\dfrac{1}{3}=\tan^2 \dfrac{\pi}{6}$

$\implies x=n\pi\pm \dfrac{\pi}{6}$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $3sin\alpha =5sin\beta ,\quad then\quad \frac { \tan { \frac { \alpha +\beta }{ 2 } } }{ \tan { \frac { \alpha -\beta }{ 2 } } } $ is equal to

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $3\sin \alpha=5\sin \beta\implies \dfrac{\sin \alpha}{\sin \beta}=\dfrac{5}{3}$


Applying componendo and dividendo rule


$\implies \dfrac{\sin \alpha+\sin \beta}{\sin \alpha-\sin \beta}=\dfrac{5+3}{5-2}$

$\implies \dfrac{2\sin \dfrac{\alpha+\beta}{2}\cos \dfrac{\alpha-\beta}{2}}{2\sin \dfrac{\alpha-\beta}{2}\cos \dfrac{\alpha+\beta}{2}}=\dfrac{8}{2}$

$\implies \dfrac{\tan \dfrac{\alpha+\beta}{2}}{\tan \dfrac{\alpha-\beta}{2}}=4$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $y\tan (A+B+C)=x\tan (A+B-C)=\lambda$, then $\tan 2C=?$

  1. $\dfrac{\lambda(x+y)}{\lambda^2-xy}$
  2. $\dfrac{\lambda(x+y)}{\lambda^2+xy}$
  3. $\dfrac{\lambda(x-y)}{xy-\lambda^2}$
  4. $\dfrac{\lambda (x-y)}{xy+\lambda^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $y\tan (A+B+C)=x\tan (A+B-C)=\lambda$

$\implies \tan (A+B+C)=\dfrac{\lambda}{y},\tan (A+B-C)=\dfrac{\lambda}{x}$

$\tan 2 C=\tan ((A+B+C)-(A+B-C))=\dfrac{\tan (A+B+C)-\tan (A+B-C)}{1+\tan (A+B+C)\tan (A+B-C)}$

                                                                                $=\dfrac{\frac{\lambda}{y}-\frac{\lambda}{x}}{1+\frac{\lambda^2}{x y}}$

                                                                                $=\dfrac{\lambda(x-y)}{x y+\lambda^2}$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $4^{2\, sin^2x}.16^{tan^2x}.2^{4\, cos^2x} = 256 $ such that $0 < x < \dfrac{\pi}{2}$ then $x$ is equal to ___________.

  1. $\dfrac{\pi}{3}$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{12}$
  4. $\dfrac{\pi}{24}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
${4}^{2{\sin}^{2}{x}}.{16}^{{\tan}^{2}{x}}.{2}^{4{\cos}^{2}{x}}=256$

$\Rightarrow\,{2}^{4{\sin}^{2}{x}}.{2}^{4{\tan}^{2}{x}}.{2}^{4{\cos}^{2}{x}}={2}^{8}$

$\Rightarrow\,{2}^{4{\sin}^{2}{x}+4{\tan}^{2}{x}+4{\cos}^{2}{x}}={2}^{8}$

$\Rightarrow\,4{\sin}^{2}{x}+4{\tan}^{2}{x}+4{\cos}^{2}{x}=8$

$\Rightarrow\,{\sin}^{2}{x}+{\tan}^{2}{x}+{\cos}^{2}{x}=2$

$\Rightarrow\,\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)+{\tan}^{2}{x}=2$

$\Rightarrow\,1+{\tan}^{2}{x}=2$ since $\left({\sin}^{2}{x}+{\cos}^{2}{x}=1\right)$

$\Rightarrow\,{\sec}^{2}{x}=2$ since $1+{\tan}^{2}{x}={\sec}^{2}{x}$

$\Rightarrow\,{\cos}^{2}{x}=\dfrac{1}{2}$

$\Rightarrow\,\cos{x}=\pm\dfrac{1}{\sqrt{2}}$

$\Rightarrow\,\cos{x}=\dfrac{1}{\sqrt{2}}$ since $0<x<\dfrac{\pi}{2}$

$\Rightarrow\,x=\dfrac{\pi}{4}$

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

The value of $\tan\left(7\dfrac{1}{2}\right)^o$ is 

  1. $\dfrac {2\sqrt 2 -(1+\sqrt 3)}{\sqrt 3-1}$
  2. $\dfrac {1+\sqrt 3}{1-\sqrt 3}$
  3. $\dfrac {1}{\sqrt 3}+\sqrt 3$
  4. $\sqrt 2 +\sqrt 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\tan\left(7\dfrac{1}{2}\right)^o=\tan\left(\dfrac{15}{2}\right)^o$


                       $=\dfrac{\sin\dfrac{15^o}{2}}{\cos\dfrac{15^o}{2}}$

                       $=\dfrac{2\sin\dfrac{15^o}{2}\times \sin\dfrac{15^o}{2}}{2\sin\dfrac{15^o}{2}\times \cos\dfrac{15^o}{2}}$

                        $=\dfrac{2\sin^2\dfrac{15^o}{2}}{\sin\left(2\times\dfrac{15^o}{2}\right)}$

                        $=\dfrac{1-\cos\left(2\times\dfrac{15^o}{2}\right)}{\sin 15^o}$

                        $=\dfrac{1-\cos 15^o}{\sin 15^o}$

Now,
$\cos 15^o=\cos(45^o-30^o)$
              $=\cos 45^o\cos30^o+\sin 45^o\sin 30^o$
              $=\dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\times\dfrac {1}{2}$
              $=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$

$\sin 15^o=\sin(45^o-30^o)$
             $=\sin 45^o\cos30^o-\cos 45^o\sin 30^o$
             $=\dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times\dfrac {1}{2}$
             $=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$


$\tan\left(7\dfrac{1}{2}\right)^o=\dfrac{1-\dfrac{\sqrt{3}+1}{2\sqrt{2}}}{\dfrac{\sqrt{3}-1}{2\sqrt{2}}}$

                      $=\dfrac{2\sqrt{2}-(\sqrt{3}+1)}{\sqrt{3}-1}$