Tag: trigonometric functions

Quizzes (5) Questions (20)

Questions Related to trigonometric functions

Multiple choice trigonometric equations trigonometric functions trigonometry maths

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

  1. $15$
  2. $45$
  3. $9$
  4. $18$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of cot^2(k*pi/11) for k=1 to 5 is a known trigonometric series result equal to 15.

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

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

  1. $1$
  2. $4$
  3. $4/\surd 3$
  4. $4\surd 3$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\tan{{5}^{\circ}}\cot{{5}^{\circ}}=1$
Multiple choice trigonometric equations trigonometric functions trigonometry maths

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

  1. $\cfrac{1}{3}$
  2. $\cfrac{1}{4}$
  3. $\cfrac{1}{9}$
  4. $\cfrac{1}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 ) $ 
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

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

  1. $- 2 / 3$
  2. $- 4 / 5$
  3. $- 3 / 5$
  4. $- 3 / 4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an isosceles right triangle, placing vertices at (0,0), (a,0), and (0,a), the medians from the acute angles are calculated. The cosine of the angle between them is -2/3.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

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

  1. $\cfrac{1}{9}$
  2. $\cfrac{1}{3}$
  3. $\cfrac{2}{3}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the altitudes of an isosceles triangle intersect on the inscribed circle, the geometry dictates that cos(A) = 1/9.

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

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