Question

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

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

Answer 1

Answer: (B)

$2\cos 2\theta$

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

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

$=\frac{sin(\theta +\frac{2\pi }{3}+\theta -\frac{\pi }{6})}{sin(\theta +\frac{2\pi }{3}-\theta +\frac{\pi }{6})}$

$=\frac{sin(2\theta +\frac{\pi }{2})}{sin\left(\frac{5\pi }{6}\right)}$

$=\frac{cos2\theta }{sin(\pi -\frac{\pi }{6})}$

$=cosec\left(\frac{\pi }{6}\right).cos2\theta$

$=2cos2\theta$.