Question

Solve:

$mtan(\theta -30)=ntan(\theta =120)$ then $\frac{m+n}{m-n}$=

• A. $cos2\theta$
• B. $2cos2\theta$
• C. $sin2\theta$
• D. $2sin2\theta$

Answer 1

The correct option is B $2cos2\theta$

Given $m\tan (\theta -30)=n\tan (\theta +120)$

$\frac{m}{n}=\frac{\tan (\theta +120)}{\tan (\theta -30)}$

Let $\theta +120=A$

$\theta -30=B$

Applying componendo dividendo

$\frac{m+n}{m-n}=\frac{\tan A+\tan B}{\tan A-\tan }$

$=\frac{\sin A\cos A+\sin B\cos A}{\sin A\cos B-\sin B\cos A}$

$=\frac{\sin (A+B)}{\sin (A-B)}$

$=\frac{\sin (\theta +120+\theta -30)}{\sin (\theta +120-\theta +30)}$

$=\frac{\sin (2\theta +90)}{\sin 150}$

$=\frac{\cos 2\theta }{1/2}$

$=2\cos 2\theta$