Tag: trigonometric functions

Questions Related to trigonometric functions

$\cot^{2} \dfrac{\pi}{11}+\cot^{2} \dfrac{2\pi}{11}+\cot^{2} \dfrac{3\pi}{11}........+\cot^{2} \dfrac{5\pi}{11}=?$

trigonometric equations trigonometric functions trigonometry maths

  1. $15$

  2. $45$

  3. $9$

  4. $18$

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Correct Option: A

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

trigonometric equations trigonometric functions trigonometry maths

  1. $\cot \alpha $

  2. $\tan \alpha $

  3. $\cos \alpha $

  4. $sin \alpha $

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Correct Option: A

Simplify: $\tan5\tan { 30 } \times 4\tan { 85=\ _ \ _ \ _  } $

trigonometric equations trigonometric functions trigonometry maths

  1. $1$

  2. $4$

  3. $4/\surd 3$

  4. $4\surd 3$

Show answer
Correct Option: C
Explanation:
$\tan{{5}^{\circ}}\tan{{30}^{\circ}}\times 4\tan{{85}^{\circ}}$
$=4\tan{{5}^{\circ}}\times\dfrac{1}{\sqrt{3}}\tan{\left({90}^{\circ}-{5}^{\circ}\right)}$
$=4\tan{{5}^{\circ}}\times\dfrac{1}{\sqrt{3}}\cot{{5}^{\circ}}$
$=\dfrac{4}{\sqrt{3}}$ since $\tan{{5}^{\circ}}\cot{{5}^{\circ}}=1$

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. $- 3 / 4$

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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 $0<x<\dfrac{\pi}{2}$

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