निम्नलिखित को सिद्ध कीजिए
$\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$
It is known that $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$
$\therefore$ $L.H.S.$ $=\tan 4 x=\tan 2(2 x)$
$=\frac{2 \tan 2 x}{1-\tan ^{2}(2 x)}$
$=\frac{2\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)}{1-\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)^{2}}$
$=\frac{\left(\frac{4 \tan x}{1-\tan ^{2} x}\right)}{\left[1-\frac{4 \tan ^{2} x}{\left(1-\tan ^{2} x\right)^{2}}\right]}$
$=\frac{\left(\frac{4 \tan x}{1-\tan ^{2} x}\right)}{\left[\frac{\left(1-\tan ^{2} x\right)^{2}-4 \tan ^{2} x}{\left(1-\tan ^{2} x\right)^{2}}\right]}$
$=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{\left(1-\tan ^{2} x\right)^{2}-4 \tan ^{2} x}$
$=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1+\tan ^{4} x-2 \tan ^{2} x-4 \tan ^{2} x}$
$=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}= R . H.S.$
माना $\cos (\alpha+\beta)=\frac{4}{5}$ और $\sin (\alpha-\beta)=\frac{5}{13},$ जहाँ $0 \leq \alpha, \beta \leq \frac{\pi}{4}$ तो $\tan 2 \alpha$ बराबर है
यदि $90^\circ < A < 180^\circ $ तथा $\sin A = \frac{4}{5},$ तब $\tan \frac{A}{2}$ का मान होगा
$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $
$2\cos x - \cos 3x - \cos 5x = $
$\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right) = $