question_answer

What is \[\frac{\cos 7x-\cos 3x}{\sin 7x-2\sin 5x+\sin 3x}\] equal to?

A) \[\tan x\]            

B) \[\cot x\]

C) \[tan2x\]          

D) \[\cot 2x\]

Correct Answer: B

Solution :

\[\frac{\cos 7x-\cos 3x}{\sin 7x-2\sin 5x+\sin 3x}\] \[=\frac{-2\sin \frac{7x+3x}{2}.\sin \frac{7x-3x}{2}}{2\sin \frac{7x+3x}{2}.\cos \frac{7x-3x}{2}-2\sin 5x}\] \[=\frac{-2\sin 5x.\sin 2x}{2\sin 5x\cos 2x-2\sin 5x}\] \[=\frac{-2\sin 5x.\sin 2x}{-2\sin 5x[1-\cos 2x]}\] \[=\frac{\sin 2x}{1-1+2{{\sin }^{2}}x}\,\,\,\,\,\,\,(\because \,\,\cos 2x=1-2{{\sin }^{2}}x)\] \[=\frac{2\sin x\cos x}{2{{\sin }^{2}}x}=\cot x\]