Question

If $\tan \theta =\frac{12}{13},$ find the value of $\frac{2\sin \theta \cos \theta }{{\cos }^2\theta -{\sin }^2\theta }$

Answer 1

Given: $\tan \theta =\frac{12}{13}$

$\frac{2\sin \theta \cos \theta }{{\cos }^2\theta -{\sin }^2\theta }=\frac{\sin 2\theta }{\cos 2\theta }=\tan 2\theta$

$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{2\times \frac{12}{13}}{1-\frac{144}{169}}=\frac{\frac{24}{13}}{\frac{25}{169}}=\frac{72}{25}$

$\frac{2\sin \theta \cos \theta }{{\cos }^2\theta -{\sin }^2\theta }=\frac{72}{25}$