Question

Show that: $\tan 3x.\tan 2x.\tan x=\tan 3x-\tan 2x-\tan x$.

Answer 1

$\Rightarrow$ Let us take, $\tan \left(3x\right)=\tan (x+2x)$

$\Rightarrow$ $\tan (A+B)=\frac{\tan \left(A\right)+\tan \left(B\right)}{1-\tan A.\tan B}$

$\tan \left(3x\right)=\frac{\tan \left(x\right)+\tan \left(2x\right)}{1-\left(\tan \right(x\left)\tan \right(2x\left)\right)}$

$\therefore$ $\tan \left(3x\right)-\tan \left(x\right).\tan \left(2x\right).\tan \left(3x\right)=\tan \left(x\right)+\tan \left(2x\right)$

$\therefore$ $\tan \left(3x\right).\tan \left(2x\right).\tan \left(x\right)=\tan \left(3x\right)-\tan \left(2x\right)-\tan \left(x\right)$