Trigonometric Identities Question 135
Question: $ \frac{\sec 8A-1}{\sec 4A-1}= $
[MP PET 1995]
Options:
A) $ \frac{\tan 2A}{\tan 8A} $
B) $ \frac{\tan 8A}{\tan 2A} $
C) $ \frac{\cot 8A}{\cot 2A} $
D) None of these
Show Answer
Answer:
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}. $
BETA
