Question

$\frac{tan5\Theta +tan3\Theta }{tan5\Theta -tan3\Theta }=4cos2\Theta cos4\Theta$

Answer 1

$\begin{array}{c}\frac{\tan 5\theta +\tan 3\theta }{\tan 5\theta -\tan 3\theta }=4\cos 2\theta \cos 4\theta \end{array}$

take $L.H.S$

$\begin{array}{c}=\frac{\frac{\sin 5\theta }{\cos 5\theta }+\frac{\sin 3\theta }{\cos 3\theta }}{\frac{\sin 5\theta }{\cos 5\theta }-\frac{\sin 3\theta }{\cos 3\theta }}\\ =\frac{\frac{\sin 5\theta \cos 3\theta +\cos 5\theta \sin 3\theta }{\sin 5\theta \cos 3\theta }}{\frac{\sin 5\theta \cos 3\theta -\cos 5\theta \sin 3\theta }{\cos 5\theta \cos 3\theta }}\\ =\frac{\sin 8\theta }{\sin 2\theta }=\frac{2\sin 4\theta \cos 4\theta }{\sin 2\theta }\\ =\frac{2\cos 4\theta \left(2\sin 2\theta \cos 2\theta \right)}{\sin 2\theta }\\ =4\cos 4\theta \cos 2\theta \end{array}$

$4\cos 2\theta \cos 4\theta$