Q. If$\ \frac{2\sin\alpha}{1 + \cos{\alpha + \sin\alpha}} = x,\ then\ \frac{1 - \cos{\alpha - \sin\alpha}}{\cos\alpha}\$is equal to
A) $\frac{1}{x}$
B) $x$
C) $1 - x$
D) None of these
Explanation
(d)
Given, $\frac{2\sin\alpha}{1 + \cos{\alpha + \sin\alpha}} = x$
$\Rightarrow \ \frac{2\sin{\alpha(1 - \cos{\alpha - \sin{\alpha)}}}}{(1 + \cos{\alpha + \sin{\alpha)\ (1 - \cos{\alpha - \sin{\alpha)}}}}} = x$
$\Rightarrow \frac{2\sin{\alpha\ (1 - \cos{\alpha - \sin{\alpha)}}}}{1 - \sin^{2}{\alpha - \cos^{2}{\alpha - 2\sin{\alpha\cos{\alpha\ }}}}} = x$
$\Rightarrow \frac{1 - \cos{\alpha - \sin\alpha}}{\cos\alpha} = - x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.