Question

If $\frac{1-\cos x}{\cos x(1+\cos x)}=\frac{\sin \alpha }{\cos x}-\frac{2}{1+\cos x}$ then $\alpha =$

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

Answer: (C)

$\frac{\pi }{2}$

Given, $\frac{1-\cos x}{\cos x(1+\cos x)}=\frac{\sin \alpha }{\cos x}-\frac{2}{1+\cos x}$ ....(1)
Resolving into partial fractions,
$\frac{1-\cos x}{\cos x(1+\cos x)}=\frac{A}{\cos x}+\frac{B}{1+\cos x}$ ....(2)
$1-\cos x=A(1+\cos x)+B\cos x$
$1-\cos x=A+(A+B)\cos x$
$\Rightarrow A=1,B=-2$
Put these values in (2)
$\frac{1-\cos x}{\cos x(1+\cos x)}=\frac{1}{\cos x}-\frac{2}{1+\cos x}$
Comparing this with (1)
$\sin \alpha =1$
$\Rightarrow \alpha =\frac{\pi }{2}$