Search any question & find its solution
Question:
Answered & Verified by Expert
If $\sin \theta=3 \sin (\theta+2 \alpha)$, then the value of $\tan (\theta+\alpha)+$
$2 \tan \alpha$ is equal to
Options:
$2 \tan \alpha$ is equal to
Solution:
1882 Upvotes
Verified Answer
The correct answer is:
$\underline{0}$
Given, $\sin \theta=3 \sin (\theta+2 \alpha)$
$\Rightarrow \frac{\sin (\theta+2 \alpha)}{\sin \theta}=\frac{1}{3}$
Apply componendo and divide do rule $\Rightarrow \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{1+3}{1-3}$
$\Rightarrow \frac{2 \sin (\theta+\alpha) \cos \alpha}{2 \cos (\theta+\alpha) \sin \alpha}=\frac{4}{-2}=-2$
$\Rightarrow \frac{\tan (\theta+\alpha)}{\tan \alpha}=-2$
$\Rightarrow \tan (\theta+\alpha)=-2 \tan \alpha \Rightarrow \tan (\theta+\alpha)+2 \tan \alpha=0$
$\Rightarrow \frac{\sin (\theta+2 \alpha)}{\sin \theta}=\frac{1}{3}$
Apply componendo and divide do rule $\Rightarrow \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{1+3}{1-3}$
$\Rightarrow \frac{2 \sin (\theta+\alpha) \cos \alpha}{2 \cos (\theta+\alpha) \sin \alpha}=\frac{4}{-2}=-2$
$\Rightarrow \frac{\tan (\theta+\alpha)}{\tan \alpha}=-2$
$\Rightarrow \tan (\theta+\alpha)=-2 \tan \alpha \Rightarrow \tan (\theta+\alpha)+2 \tan \alpha=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.