Question

The value of $\tan A\sec 4A+\tan 4A-\tan A-\tan 4A\sec 2A=$

Answer 1

$\tan A\sec 4A+\tan 4A-\tan A-\tan 4A\sec 2A=\tan A(\sec 4A-1)+\tan 4A(1-\sec 2A)$

$=\tan A\left(\frac{1-\cos 4A}{\cos 4A}\right)+\tan 4A\frac{(\cos 2A-1)}{\cos 2A}$

$=\frac{\sin A}{\cos A}\left(\frac{2{\sin }^{2}2A}{\cos 4A}\right)-\frac{2\sin 4A}{\cos 4A}\frac{\left({\sin }^{2}A\right)}{\cos 2A}$

$=\frac{2\sin A{\sin }^{2}2A}{\cos A\cos 4A}-\frac{2\left(2\sin 2A{\cos }^{2}A\right)\sin 2A}{\cos 4A\cos 2A}$

$=\frac{2\sin A.4{\sin }^{2}A\cos A}{\cos 4A}-\frac{4\left(2\sin A\cos A\right){\sin }^{2}A}{\cos 4A}$

$=\frac{8{\sin }^{3}A\cos A}{\cos 4A}-\frac{8{\sin }^{3}A\cos A}{\cos 4A}=0$