Question

Prove $\tan \left(3\theta \right)=\frac{(3\tan \theta -{\tan }^3\theta )}{(1-3{\tan }^2\theta )}$

Answer 1

Prove that : $\tan \left(3\theta \right)=\frac{(3\tan \theta -{\tan }^3\theta )}{(1-3{\tan }^2\theta )}$

Use formula:

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

Solve the L.H.S part.

$$\begin{gathered} \tan \left(3\theta \right)=\tan (\theta +2\theta ) \\ \Rightarrow =\frac{\tan \theta +\tan 2\theta }{1-\tan \theta \times \tan 2\theta } \\ \Rightarrow =\frac{\tan \theta +\tan (\theta +\theta )}{1-\tan \theta \times \tan (\theta +\theta )}\because \tan 2\theta =\tan (\theta +\theta ) \\ \Rightarrow =\frac{\tan \theta +{\displaystyle \frac{\tan \theta +\tan \theta }{1-{\tan }^2\theta }}}{1-{\displaystyle \frac{\tan \theta (\tan \theta +\tan \theta )}{1-{\tan }^2\theta }}} \\ \Rightarrow =\frac{{\displaystyle \frac{\tan \theta (1-{\tan }^2\theta )+\tan \theta +\tan \theta }{1-{\tan }^2\theta }}}{{\displaystyle \frac{(1-{\tan }^2\theta )-\tan \theta (\tan \theta +\tan \theta )}{1-{\tan }^2\theta }}} \\ \Rightarrow =\frac{{\displaystyle \frac{\tan \theta -{\tan }^3\theta +\tan \theta +\tan \theta }{1-{\tan }^2\theta }}}{{\displaystyle \frac{1-{\tan }^2\theta -{\tan }^2\theta -{\tan }^2\theta }{1-{\tan }^2\theta }}} \\ \Rightarrow =\frac{{\displaystyle \frac{3\tan \theta -{\tan }^3\theta }{1-{\tan }^2\theta }}}{{\displaystyle \frac{1-3{\tan }^2\theta }{1-{\tan }^2\theta }}} \\ \Rightarrow =\frac{3\tan \theta -{\tan }^3\theta }{1-{\tan }^2\theta }\times \frac{1-{\tan }^2\theta }{1-3{\tan }^2\theta } \\ \Rightarrow =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta } \end{gathered}$$

Hence, the L.H.S= R.H.S.