Tag: trigonometry

Questions Related to trigonometry

$tan  5x-tan  3x-tan  2x=$

  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 $\tan{{5}^{\circ}}\cot{{5}^{\circ}}=1$