Show that
$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
Show that
$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
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
If $\alpha$, $\beta$,$\gamma$ are positive number such that $\alpha + \beta = \pi$ and $\beta + \gamma = \alpha$, then $tan\ \alpha$ is equal to - (where $\gamma \ne n\pi ,n \in I$ )
If $\alpha$, $\beta$,$\gamma$ are positive number such that $\alpha + \beta = \pi$ and $\beta + \gamma = \alpha$, then $tan\ \alpha$ is equal to - (where $\gamma \ne n\pi ,n \in I$ )
$\cos \alpha .\sin (\beta - \gamma ) + \cos \beta .\sin (\gamma - \alpha ) + \cos \gamma .\sin (\alpha - \beta ) = $
$\cos \alpha .\sin (\beta - \gamma ) + \cos \beta .\sin (\gamma - \alpha ) + \cos \gamma .\sin (\alpha - \beta ) = $
If $A + B + C = \pi ,$ then ${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + $${\tan ^2}\frac{C}{2}$ is always
If $A + B + C = \pi ,$ then ${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + $${\tan ^2}\frac{C}{2}$ is always
If $x = \cos 10^\circ \cos 20^\circ \cos 40^\circ ,$ then the value of $x$ is
If $x = \cos 10^\circ \cos 20^\circ \cos 40^\circ ,$ then the value of $x$ is
If $\cos A = \frac{3}{4}$, then $32\sin \frac{A}{2}\cos \frac{5}{2}A = $
If $\cos A = \frac{3}{4}$, then $32\sin \frac{A}{2}\cos \frac{5}{2}A = $