Question

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

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

Answer 1

Answer: (A)

By the given conditions
$\tan \frac{\pi }{9}+\tan \frac{5\pi }{18}=2x;\tan \frac{\pi }{9}+\tan \frac{7\pi }{18}=2y$
$\Rightarrow 2x=\tan 20^{\circ }+\tan 50^{\circ }$
$=\frac{\sin 20^{\circ }}{\cos 20^{\circ }}+\frac{\sin 50^{\circ }}{\cos 50^{\circ }}=\frac{\sin 20^{\circ }\cos 50^{\circ }+\cos 20^{\circ }\sin 50^{\circ }}{\cos 20^{\circ }\cos 50^{\circ }}$
$=\frac{\sin 70^{\circ }}{\cos 20^{\circ }\cos 50^{\circ }}=\frac{\cos 20^{\circ }}{\cos 20^{\circ }\cos 50^{\circ }}$
$=\frac{1}{\cos 50^{\circ }}=\frac{1}{\sin 40^{\circ }}=\mathrm{cosec}40^{\circ }$
and $2y=\tan 20^{\circ }\tan 70^{\circ }$
$=\frac{\sin 20^{\circ }}{\cos 20^{\circ }}+\frac{\sin 70^{\circ }}{\cos 70^{\circ }}=\frac{\sin 90^{\circ }}{\cos 20^{\circ }\cos 70^{\circ }}$
$=\frac{1}{\cos 20^{\circ }\cos 70^{\circ }}=\frac{1}{\cos 20^{\circ }\sin 20^{\circ }}$
$=\frac{2}{2\sin 20^{\circ }\cos 20^{\circ }}=\frac{2}{\sin 40^{\circ }}=2\mathrm{cosec}40^{\circ }$
$\therefore 2y=2(2x)$ or $y=2x$