Question

Show that $tan3\theta =\frac{3tan\theta -ta{n}^3\theta }{1-3ta{n}^2\theta }$

Answer 1

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

$\tan (2\theta +\theta )=\frac{\tan 2\theta +\tan \theta }{1-\tan \theta \tan 2\theta }$

$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

$=\frac{\frac{2\tan \theta }{1-{\tan }^2\theta }+\tan \theta }{1-\tan \theta \left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)}$

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

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

Hence proof