Question

If $\tan \frac{\pi }{9},x$ and $\tan \frac{5\pi }{18}$are in AP and $\tan \frac{\pi }{9},y$ and $\tan \frac{7\pi }{18}$ are also in AP then

• A. $2x=y$
• B. $x>y$
• C. $x=y$
• D. none of these

Answer 1

The correct option is A $2x=y$

$\because \tan \frac{\pi }{9},x\&\tan \frac{5\pi }{18}\Rightarrow 2x=\tan \frac{\pi }{9}+\tan \frac{5\pi }{18}\Rightarrow 2x=\frac{\sin (\frac{\pi }{9}+\frac{5\pi }{18})}{\cos \frac{\pi }{9}\cos \frac{5\pi }{18}}=\frac{\sin \frac{7\pi }{18}}{\sin (\frac{\pi }{2}-\frac{\pi }{9})\sin (\frac{\pi }{2}-\frac{5\pi }{18})}$
$\Rightarrow 2x=\frac{\sin \frac{7\pi }{18}}{\sin \frac{7\pi }{18}\sin \frac{2\pi }{9}}=\csc \frac{2\pi }{9}$ ...(1)
$\because \tan \frac{\pi }{9},y\&\tan \frac{7\pi }{18}\Rightarrow 2y=\tan \frac{\pi }{9}+\tan \frac{7\pi }{18}\Rightarrow 2y=\frac{\sin (\frac{\pi }{9}+\frac{7\pi }{18})}{\cos \frac{\pi }{9}\cos \frac{7\pi }{18}}=\frac{\sin \frac{\pi }{2}}{\cos \frac{\pi }{9}\sin (\frac{\pi }{2}-\frac{7\pi }{18})}$
$\Rightarrow 2y=\frac{2}{2\cos \frac{\pi }{9}\sin \frac{\pi }{9}}=\frac{2}{\sin \frac{\pi }{9}}=2\csc \frac{2\pi }{9}$ ..(2)
From (1) and (2)
$2x=y$
Ans: A