Tag: trigonometry

Questions Related to trigonometry

If $\alpha$ is the angle of first quadrant such that $co\sec ^{ 4 }{ \alpha  }=17+\cot ^{ 4 }{ \alpha  } $, then what is the value of $\sin{\alpha}$?

trigonometric equations trigonometric functions trigonometry maths

  1. $\cfrac{1}{3}$

  2. $\cfrac{1}{4}$

  3. $\cfrac{1}{9}$

  4. $\cfrac{1}{16}$

Show answer
Correct Option: A
Explanation:
$ cosec^{4}\alpha -cot^{4}\alpha = 17$

 (As $ cosec^{2}\alpha -cot^{2}\alpha=1) $

$ \Rightarrow (cosec^{2}\alpha -cot^{2}\alpha )(cosec^{2}\alpha +cot^{2}\alpha ) = 17 $ 

$ \Rightarrow cosec^{2}\alpha +cot^{2}\alpha = 17...(1) $

$ cosec^{2}\alpha -cot^{2}\alpha = 1...(2) $

then $ (1) + (2) \Rightarrow 2cosec^{2}\alpha = 18 $

$ \Rightarrow sin^{2}\alpha = \dfrac{1}{9}\Rightarrow \boxed{sin\,\alpha = \dfrac{1}{3}} $ $ \left ( \because \alpha \,in\,1st\,quadrant \right ) $ 

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

trigonometric equations trigonometric functions trigonometry maths

  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. $2n\pi\pm\dfrac{\pi}{6},n\in Z$

Show answer
Correct Option: C
Explanation:

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


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


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

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

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

The cosine of the obtuse angle formed by the medians from the vertices of the acute angles of an isosceles right angled triangle is

trigonometric equations trigonometric functions trigonometry maths

  1. $- 2 / 3$

  2. $- 4 / 5$

  3. $- 3 / 5$

  4. $- 3 / 4$

Show answer
Correct Option: A

In an isosceles $\triangle ABC$, if the altitudes intersect on the inscribed circle then cosine of the vertical angle $'A'$ is :

trigonometric equations trigonometric functions trigonometry maths

  1. $\cfrac{1}{9}$

  2. $\cfrac{1}{3}$

  3. $\cfrac{2}{3}$

  4. None of these

Show answer
Correct Option: A

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

trigonometric equations trigonometric functions trigonometry maths

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: A
Explanation:

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


Applying componendo and dividendo rule


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

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

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

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

trigonometric equations trigonometric functions trigonometry maths

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

Show answer
Correct Option: B
Explanation:

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

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

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

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

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

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 ___________.

trigonometric equations trigonometric functions trigonometry maths

  1. $\dfrac{\pi}{3}$

  2. $\dfrac{\pi}{4}$

  3. $\dfrac{\pi}{12}$

  4. $\dfrac{\pi}{24}$

Show answer
Correct Option: B
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 $0<x<\dfrac{\pi}{2}$

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

The value of ${ cosec }^{ 2 }{ 51 }^{ 0 }-{ cot }^{ 2 }{ 51 }^{ 0 }$ is 

using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

  1. 1

  2. -1

  3. 0

  4. $\frac { 1 }{ 2 } $

Show answer
Correct Option: A
Explanation:

We know,

$1 + \cot^2\theta = cosec^2 \theta$

$\therefore cosec^2\theta - \cot^2 \theta = 1$

Thus, $cosec^2  51^o - \cot^251^o = 1$

Hence, option a is correct.

Choose the correct option for the following statement.

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

  1. The given statement is true 

  2. The given statement is false

  3. Incomplete information

  4. None of the above.

Show answer
Correct Option: A
Explanation:

By definition,

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
Therefore, the given statement is true.

$\sin { { 30 }^{ o }= } $

using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

  1. $\dfrac {1}{2}$

  2. $-\dfrac {1}{2}$

  3. $\dfrac {\sqrt {3}}{2}$

  4. $-\dfrac {\sqrt {3}}{2}$

Show answer
Correct Option: A
Explanation:

As we know that 

$\sin 30^{\circ}=\dfrac{1}{2}$