Question

If $\tan \theta =\frac{12}{13}$, evaluate $\frac{2\sin \theta \cos \theta }{{\cos }^2\theta -{\sin }^2\theta }$

Answer 1

$\begin{array}{c}\tan \theta =\frac{12}{13}\\ \Rightarrow \frac{2\sin \theta .\cos \theta }{{\cos }^2\theta -{\sin }^2\theta }=\frac{\sin 2\theta }{\cos 2\theta }\\ \Rightarrow \tan 2\theta \left[\begin{array}{c}2\sin \theta .\cos \theta =\sin 2\theta \\ {\cos }^{2}x-{\sin }^{2}x=\cos 2x\end{array}\right]\\ \Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{\frac{24}{13}}{1-\frac{144}{169}}=\frac{24}{13}\times \frac{169}{25}\\ \tan 2\theta =\frac{312}{25}Ans.\end{array}$