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
$\tan 5x\tan 3x\tan 2x = $
$\tan 5x\tan 3x\tan 2x = $
$\cos 15^\circ - \sin 15^\circ $ का मान है
$\cos 15^\circ - \sin 15^\circ $ का मान है
$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}$ बराबर है
$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}$ बराबर है
- [JEE MAIN 2023]
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
- [IIT 1988]
यदि $A + B + C = \pi ,$ तो ${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + $${\tan ^2}\frac{C}{2}$ हमेशा है
यदि $A + B + C = \pi ,$ तो ${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + $${\tan ^2}\frac{C}{2}$ हमेशा है