Question

Prove the following :
$\tan 8\theta -\tan 5\theta -\tan 3\theta =\tan 8\theta .\tan 5\theta .\tan 3\theta$

Answer 1

We know that,

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

$\therefore L.H.S=\tan 8\theta -\tan 5\theta -\tan 3\theta =\tan 8\theta -(\tan 5\theta +\tan 3\theta )=\tan 8\theta -\left[\tan (5\theta +3\theta )(1-\tan 5\theta \tan 3\theta )\right]=\tan 8\theta -\left[\tan 8\theta (1-\tan 5\theta \tan 3\theta )\right]=\tan 8\theta -(\tan 8\theta -\tan 8\theta \tan 5\theta \tan 3\theta )=\tan 8\theta -\tan 8\theta +\tan 8\theta \tan 5\theta \tan 3\theta =0+\tan 8\theta \tan 5\theta \tan 3\theta =\tan 8\theta \tan 5\theta \tan 3\theta =R.H.S$

Hence, proved.