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  4. If tan (π/9), x and tan (5 π/18) are in A.P. and tan (π/9), y and tan (7 π/18) are also in A.P. then

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Q. 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

Trigonometric Functions

Solution:

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