0

Trigonometric functions - class-XI

Description: trigonometric functions
Number of Questions: 20
Created by:
Tags: trigonometric functions maths trigonometry
Attempted 0/20 Correct 0 Score 0

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

  1. $1$

  2. $-1$

  3. $ \pm 1$

  4. $2$


Correct Option: B
Explanation:

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

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

  1. True

  2. False


Correct Option: B
Explanation:

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

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

State true or false

$\tan{ 18 }^{ 0 }+\tan27^{ 0 }+\tan{ 18 }^{ 0 }\cdot \tan27^{ 0 }=1$

  1. True

  2. False


Correct Option: B
Explanation:

$tan180+tan270+tan180+tan270=1$

$0.32491969 + 0.50952544+0.32491969 + 0.50952544 = 1$

$1.66889026 = 1$

The left side $1.66889026 $ not equal to rightside$ 1$.

Which means the given statement is false.

 

$tan  5x-tan  3x-tan  2x=$

  1. $\tan 5x \tan 3x \tan 2x$

  2. $\sin 5x \sin 3x \sin 2x$

  3. $\cos 5x \cos 3x \cos 2x$

  4. $\sec 5x \sec 3x \sec 2x$


Correct Option: A
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, $\tan 5x-\tan 3x-\tan 2x=\tan 2x.\tan 3x.\tan 5x$.

$A, B, C$ are three angles such that $\tan  A+\tan  B+\tan  C=\tan  A  \tan  B  \tan  C.$ Which of the following statements is always correct ?

  1. $ABC$ is a triangle, i.e. $A+B+C=\pi $

  2. $A=B=C. i.e., $ $ABC$ is an equilateral triangle

  3. $A+B=C, $ i.e., $ABC$ is a right- angled triangle

  4. $A+B=\pi $


Correct Option: A
Explanation:

(A) $tan\left [ (A+B)+C \right ]$
$=\frac{tan (A+B)+tan C}{1-tan (A+B)  tan  C}=\frac{\frac{tan  A+tan   B}{1-tan  A   tan  B}+tan  C}{1-\frac{tan  A+tan  B}{1-tan   A   tan  B}.  tan  C}$
$=\frac{tan  A+tan  B+tan  C-tan  A   tan  B   tan  C}{Denominator}$
$=0$
$\left [ since,  tan  A+tan  B+tan  C =tan  A   tan  B   tan  C \right ]$
$\therefore A+B+C=\pi $ i.e.,  A, B, C is a triangle

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. $\pi/4$


Correct Option: A

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 } $


Correct Option: B
Explanation:

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

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

$\alpha, \beta$ are the solution (s) of $3 cos 2 \theta + 4 sin 2 \theta = 5$
$tan (\alpha + \beta) = $

  1. $1$

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

  3. $\dfrac{4}{3}$

  4. $\dfrac{1}{4}$


Correct Option: A

$\alpha, \beta$ are the solution (s) of $3 cos 2 \theta + 4 sin 2 \theta = 5$
$tan (\alpha - \beta) = $

  1. $0$

  2. $1$

  3. $\dfrac{1}{4}$

  4. $\dfrac{4}{3}$


Correct Option: A

In $\Delta$ ABC, (a + b + c) ( tan  $\dfrac{A}{2}$ + tan $\dfrac{B}{2}$) =

  1. 2 c cot $\dfrac{A}{2}$

  2. 2 c cot $\dfrac{B}{2}$

  3. 2 c cot $\dfrac{C}{2}$

  4. 2 c tan $\dfrac{C}{2}$


Correct Option: A

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

  1. $15$

  2. $45$

  3. $9$

  4. $18$


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

  1. $\cot \alpha $

  2. $\tan \alpha $

  3. $\cos \alpha $

  4. $sin \alpha $


Correct Option: A

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

  1. $1$

  2. $4$

  3. $4/\surd 3$

  4. $4\surd 3$


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 $\sin{\alpha}$?

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

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

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

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


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

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


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

  1. $- 2 / 3$

  2. $- 4 / 5$

  3. $- 3 / 5$

  4. $- 3 / 4$


Correct Option: A

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


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

  1. $1$

  2. $2$

  3. $3$

  4. $4$


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 $\tan 2C=?$

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


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 ___________.

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

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

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

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


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

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

- Hide questions