Question

Find the value of $m$ for the given question $\frac{\sec 8\theta -1}{\sec 4\theta -1}=m\frac{\tan 8\theta }{\tan 2\theta }$

Answer 1

$\frac{\sec 8\theta -1}{\sec 4\theta -1}=\frac{\tan 8\theta }{\tan 2\theta }$

$\Rightarrow L.H.S=\frac{\sec 8\theta -1}{\sec 4\theta -1}=\frac{\frac{1}{\cos 8\theta }-1}{\frac{1}{\cos 4\theta }-1}=\frac{\frac{1-\cos 8\theta }{\cos 8\theta }}{\frac{1-\cos 4\theta }{\cos 4\theta }}$

$=\frac{2{\sin }^{2}4\theta }{\cos 8\theta }\times \frac{\cos 4\theta }{2{\sin }^{2}2\theta }$

$[\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=\frac{\left(2\sin 4\theta .\cos 4\theta \right)}{\cos 8\theta }.\frac{\sin 4\theta }{2{\sin }^{2}2\theta }$

$=\frac{\sin 2\left(4\theta \right)}{\cos 8\theta }.\frac{2\sin 2\theta .\cos 2\theta }{2{\sin }^{2}2\theta }$

$[\because 2\sin \theta .\cos \theta =\sin 2\theta ]$

$=\frac{\sin 8\theta }{\cos 8\theta }.\frac{\cos 2\theta }{\sin 2\theta }$

$=\tan 8\theta .\cot 2\theta$

$=\frac{\tan 8\theta }{\tan 2\theta }$

$\therefore m=1$