Trigonometric Identities Question 89
Question: $ \frac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x} $ is equal to
Options:
A) $ \sin 2x $
B) $ \cos 2x $
C) $ \tan 2x $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}=\frac{\sin 3x-\sin x}{{{\cos }^{2}}x-{{\sin }^{2}}x} $ $ =\frac{2\cos \frac{3x+x}{2}.\sin \frac{3x-x}{2}}{\cos 2x}=\frac{2\cos 2x.\sin x}{\cos 2x} $ 0 $ =2\sin x $
BETA
