Question

Prove $\frac{{\tan }^{2}2\theta +{\tan }^2\theta }{1-{\tan }^{2}2\theta {\tan }^2\theta }=\tan 3\theta \tan \theta$.

Answer 1

L.H.S $\frac{{\tan }^{2}2\theta +{\tan }^2\theta }{1-{\tan }^{2}2\theta {\tan }^2\theta }=\frac{{\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)}^2+{\tan }^2\theta }{1-{\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)}^2{\tan }^2\theta }$
$=\frac{4{\tan }^2\theta +{\tan }^2\theta {(1-{\tan }^2\theta )}^2}{{(1-{\tan }^2\theta )}^2-\left(4{\tan }^2\theta \right){\tan }^2\theta }$
$=\frac{4{\tan }^2\theta +{\tan }^2\theta (1+{\tan }^4\theta -2{\tan }^2\theta )}{(1+{\tan }^4\theta -2{\tan }^2\theta )-4{\tan }^4\theta }$
$=\frac{4{\tan }^2\theta +{\tan }^2\theta +{\tan }^6\theta +2{\tan }^4\theta }{1+{\tan }^4\theta -2{\tan }^2\theta -4{\tan }^4\theta }$
$=\frac{{\tan }^6\theta -2{\tan }^4\theta +5{\tan }^2\theta }{1-2{\tan }^2\theta -3{\tan }^4\theta }$
R.H.S $\tan 3\theta \tan \theta =\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)\tan \theta =\frac{3{\tan }^2\theta -{\tan }^4\theta }{1-3{\tan }^2\theta }$