Question

Prove that

$\left(\tan 4A+\tan 2A\right)\left(1-{\tan }^{2}3A{\tan }^{2}A\right)=2\tan 3A{\sec }^{2}A$

Answer 1

$LHS=(\tan 4A+\tan 2A)(1-\tan {}^{2}3A\tan {}^{2}A)=(\tan 4A+\tan 2A)(1+\tan 3A\tan A)(1-\tan 3A\tan A)=\left(\frac{\sin 4A\cos 2A+\cos 4A\sin 2A}{\cos 4A\cos 2A}\right)\left(\frac{\cos 3A\cos A-\cos 3A\sin A}{\cos 3A\cos A}\right)\left(\frac{\cos 2A}{\cos 3A\cos A}\right)=\left(\frac{\sin 6A}{\cos 4A\cos 2A}\right)\left(\frac{\cos 4A}{\cos 3A\cos A}\right)\left(\frac{\cos 2A}{\cos 3A\cos A}\right)=\frac{\sin 6A}{\cos {}^{3}3A\cos {}^{2}A}=\frac{2\sin 3A\cos 3A}{\cos {}^{2}3A\cos {}^{2}A}=2\tan 3A\sec {}^{2}A=RHS$