A) \[\frac{\tan 2A}{\tan 8A}\]
B) \[\frac{\tan 8A}{\tan 2A}\]
C) \[\frac{\cot 8A}{\cot 2A}\]
D) None of these
Correct Answer: B
Solution :
\[\frac{\sec 8A-1}{\sec 4A-1}=\frac{1-\cos 8A}{\cos 8A}.\frac{\cos 4A}{1-\cos 4A}\] \[=\frac{2{{\sin }^{2}}4A}{\cos 8A}\frac{\cos 4A}{2{{\sin }^{2}}2A}\]\[=\frac{2\sin 4A\cos 4A}{\cos 8A}\frac{\sin 4A}{2{{\sin }^{2}}2A}\] \[=\tan 8A\frac{2\sin 2A\cos 2A}{2{{\sin }^{2}}2A}=\frac{\tan 8A}{\tan 2A}.\]
You need to login to perform this action.
You will be redirected in
3 sec
