Question

Prove that: $\tan 3A\tan 2A\tan A=\tan 3A-\tan 2A-\tan A$

Answer 1

Given: $\tan 3A\tan 2A\tan A=\tan 3A-\tan 2A-\tan A$

Rearranging the given relation, we have to prove that

$\Rightarrow$$\tan A+\tan 2A=\tan 3A(1-\tan A\tan 2A)$

$\Rightarrow$$\frac{\tan A+\tan 2A}{1-\tan A\tan 2A}=\tan 3A$

$\Rightarrow$$\tan (A+2A)=\tan 3A$ ......which is true

Hence proved.