Trigonometric Identities Question 244
Question: What is $ \frac{\cos 7x-\cos 3x}{\sin 7x-2\sin 5x+\sin 3x} $ equal to?
Options:
A) $ \tan x $
B) $ \cot x $
C) $ tan2x $
D) $ \cot 2x $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{\cos 7x-\cos 3x}{\sin 7x-2\sin 5x+\sin 3x} $ $ =\frac{-2\sin \frac{7x+3x}{2}.\sin \frac{7x-3x}{2}}{2\sin \frac{7x+3x}{2}.\cos \frac{7x-3x}{2}-2\sin 5x} $ $ =\frac{-2\sin 5x.\sin 2x}{2\sin 5x\cos 2x-2\sin 5x} $ $ =\frac{-2\sin 5x.\sin 2x}{-2\sin 5x[1-\cos 2x]} $ $ =\frac{\sin 2x}{1-1+2{{\sin }^{2}}x},(\because \cos 2x=1-2{{\sin }^{2}}x) $ $ =\frac{2\sin x\cos x}{2{{\sin }^{2}}x}=\cot x $
BETA
