Tag: trigonometric functions

Questions Related to trigonometric functions

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

trigonometric equations trigonometric functions trigonometry maths

  1. $1$

  2. $-1$

  3. $ \pm 1$

  4. $2$

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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 $\tan(\dfrac{\pi}{4} + \theta) - \tan(\dfrac{\pi}{4} -\theta) = 2\tan\theta$

trigonometric equations trigonometric functions trigonometry maths

  1. True

  2. False

Show answer
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

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

trigonometric equations trigonometric functions trigonometry maths

  1. True

  2. False

Show answer
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.

 

$tan  5x-tan  3x-tan  2x=$

trigonometric equations trigonometric functions trigonometry maths

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

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

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

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

Show answer
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 ?

trigonometric equations trigonometric functions trigonometry maths

  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 $

Show answer
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 $\tan^{-1}\left(\dfrac{1}{2}\tan 2A\right)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3}A)$=

trigonometric equations trigonometric functions trigonometry maths

  1. $0$

  2. $\pi$

  3. $\pi/2$

  4. $\pi/4$

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Correct Option: A

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

trigonometric equations trigonometric functions trigonometry maths

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

  2. $2$

  3. $1$

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

Show answer
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) = $

trigonometric equations trigonometric functions trigonometry maths

  1. $1$

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

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

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

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Correct Option: A

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

trigonometric equations trigonometric functions trigonometry maths

  1. $0$

  2. $1$

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

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

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Correct Option: A

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

trigonometric equations trigonometric functions trigonometry maths

  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 $\dfrac{C}{2}$

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Correct Option: A