Question

$x\tan (\theta -\frac{\pi }{6})=y\tan (\theta +\frac{2\pi }{3})⟹\frac{x+y}{x-y}=$

• A. $\cos 2\theta$
• B. $2\cos 2\theta$
• C. $\sin 2\theta$
• D. $2\sin 2\theta$

Answer 1

Answer: (B)

$2\cos 2\theta$

Given that

$x\tan (\theta -\frac{\pi }{6})=y\tan (\theta +\frac{2\pi }{3})$

$\Rightarrow x\left[\frac{\tan \theta -\tan \frac{\pi }{6}}{1+\tan \theta \cdot \tan \frac{\pi }{6}}\right]=y\left[\frac{\tan \theta +\tan \frac{2\pi }{3}}{1-\tan \theta \cdot \tan \frac{2\pi }{3}}\right]$

$\Rightarrow x\left[\frac{\tan \theta -1/\sqrt{3}}{1+\frac{1}{\sqrt{3}}\cdot \tan \theta }\right]=y\left[\frac{\tan \theta +\left(\sqrt{3}\right)}{1-\sqrt{3}\tan \theta }\right]$

$\Rightarrow x\left[\frac{\sqrt{3}\tan \theta -1}{\sqrt{3}+\tan \theta }\right]=y\left[\frac{\tan \theta +\left(\sqrt{3}\right)}{1+\sqrt{3}\tan \theta }\right]$

$\Rightarrow x[3{\tan }^2\theta -1]=y[{\tan }^2\theta -3]$

$\Rightarrow \frac{x}{y}=\left[\frac{{\tan }^2\theta -3}{3{\tan }^2\theta -1}\right]\text{(Apply componendo-dividendo)}$

$\Rightarrow \frac{x+y}{x-y}=\frac{{\tan }^2\theta -3+3{\tan }^2\theta -1}{{\tan }^2\theta -3-3{\tan }^2\theta +1}$

$\Rightarrow \frac{x+y}{x-y}=\frac{4{\tan }^2\theta -4}{-2{\tan }^2\theta -2}=\frac{-4(1-{\tan }^2\theta )}{-2({\tan }^2\theta +1)}$

$\Rightarrow \frac{x+y}{x-y}=2\left(\frac{1-\frac{{\sin }^2\theta }{{\cos }^2\theta }}{\frac{{\sin }^2\theta }{{\cos }^2\theta }+1}\right)=2\left[\frac{{\cos }^2\theta -{\sin }^2\theta }{{\sin }^2\theta +{\cos }^2\theta }\right]$

$\Rightarrow \frac{x+y}{x-y}=2[{\cos }^2\theta -{\sin }^2\theta ]$

$\Rightarrow \frac{x+y}{x-y}=2\cos 2\theta$

$\text{hence,}\left(B\right)\text{is the correct option}$