Question

❤❤❤Hello frds❤❤❤


☺☺plz answer my question ☺☺



$\frac{ \sec(8 \alpha ) \: - 1 }{ \sec(4 \alpha ) \: - 1 } = \frac{ \tan(8 \alpha ) }{ \tan( 2 \alpha ) }$


All the best ✌✌✌

Answer 1

Given,

$\frac{\sec8\alpha\;-\;1}{\sec4\alpha\;-\;1}\\\;\\=\frac{\frac{1}{\cos8\alpha}-1}{\frac{1}{\cos4\alpha}-1}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{1-\cos8\alpha}{1-\cos4\alpha}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{1- 1 +2\sin^{2}4\alpha}{1- 1 +2\sin^{2}2\alpha}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{2\sin^{2}4\alpha}{2\sin^{2}2\alpha}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{\sin^{2}4\alpha}{\sin^{2}2\alpha}\\\;\\=\frac{2\sin4\alpha.\cos4\alpha.2\sin2\alpha.\cos2\alpha}{2\cos8\alpha.\sin^{2}2\alpha}\\\;\\=\frac{\sin8\alpha.\cos2\alpha}{\cos8\alpha.sin2\alpha}\\\;\\=\frac{tan8\alpha}{tan2\alpha}\\\;\\Note:1. cos2\theta= 1-2\sin^{2}\theta\\\;\\\sin2A=2\sin A.\cos A$