दिखाइए
$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
दिखाइए
$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
We know that $3 x=2 x+x$
Therefore, $\tan 3 x=\tan (2 x+x)$
or $\tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}$
or $\tan 3 x-\tan 3 x \tan 2 x \tan x=\tan 2 x+\tan x$
or $\tan 3 x-\tan 2 x-\tan x=\tan 3 x \tan 2 x \tan x$
or $\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
Similar Questions
यदि $2\tan A = 3\tan B,$ तब $\frac{{\sin 2B}}{{5 - \cos 2B}}$ का मान होगा
यदि $2\tan A = 3\tan B,$ तब $\frac{{\sin 2B}}{{5 - \cos 2B}}$ का मान होगा
यदि $A + B + C = {180^o},$ तो $\frac{{\tan A + \tan B + \tan C}}{{\tan A\,.\,\tan B\,.\,\tan C}} = $
यदि $A + B + C = {180^o},$ तो $\frac{{\tan A + \tan B + \tan C}}{{\tan A\,.\,\tan B\,.\,\tan C}} = $
$2\cos x - \cos 3x - \cos 5x = $
$2\cos x - \cos 3x - \cos 5x = $
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
- [IIT 1986]
$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $
$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $