Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{{\tan }^{2}2\theta -{\tan }^2\theta }{1-{\tan }^{2}2\theta {\tan }^2\theta } \\ =\frac{(\tan 2\theta +\tan \theta )(\tan 2\theta -\tan \theta )}{1-{\tan }^{2}2\theta {\tan }^2\theta }\left\{\mathrm{Using}{A}^2-B^2=\left(A+B\right)\left(A-B\right)\right\} \\ \tan 3\theta =\tan (2\theta +\theta )\text{a}\text{n}\text{d}\tan \theta =\tan (2\theta -\theta ). \\ =\frac{\tan 3\theta \left(1-\tan 2\theta \tan \theta \right)\times \tan \theta \left(1+\tan 2\theta \tan \theta \right)}{1-{\tan }^{2}2\theta {\tan }^2\theta }\left[\because \tan 2\theta +\tan \theta =\tan 3\theta \left(1-\tan 2\theta \tan \theta \right)\&\tan 2\theta -\tan \theta =\tan \theta \left(1+\tan 2\theta \tan \theta \right)\right] \\ =\frac{\tan 3\theta \tan \theta (1-{\tan }^{2}2\theta {\tan }^2\theta )}{1-{\tan }^{2}2\theta {\tan }^2\theta } \\ =\tan 3\theta \tan \theta \end{gathered}$$

$$\begin{gathered} =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$