Question

What is $\frac{\sin 5x-\sin 3x}{\cos 5x+\cos 3x}$ equal to?

• A. $\sin x$
• B. $\cos x$
• C. $\tan x$
• D. $\cot x$

Answer 1

Answer: (C)

$\tan x$

$\frac{\sin 5x-\sin 3x}{\cos 5x+\cos 3x}$

$\to \sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

$\to \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$\therefore \sin 5x-\sin 3x=2\cos 4x\sin x$

$\cos 5x+\cos 3x=2\cos 4x\cos x$

$\frac{\sin 5x-\sin 3x}{\cos 5x+\cos 3x}=\frac{\sin x}{\cos x}=\tan x$.