Q. If $\theta = \frac{2\sin x}{1 + \sin{x + \cos x}}$, then $\frac{1 + \sin{x - \cos x}}{1 + \sin x}$ equals

A) 0
B) $- \theta$
C) $\theta$
D) $- \theta/2$
Explanation

(c)
Given, $\theta = \frac{2\sin x}{1 + \sin{x + \cos x}}$
$\Rightarrow \ \ \theta = \frac{4\sin{\frac{x}{2}\cos\frac{x}{2}}}{2\cos^{2}{\frac{x}{2} + 2\sin{\frac{x}{2}\cos\frac{x}{2}}}}$
$\Rightarrow \ \ \theta = \frac{2\sin\frac{x}{2}}{\cos{\frac{x}{2} + \sin\frac{x}{2}}} \times \frac{\left( \sin{\frac{x}{2} + \cos\frac{x}{2}} \right)}{\left( \sin{\frac{x}{2} + \cos\frac{x}{2}} \right)}$
$\Rightarrow \ \ \theta = \frac{1 - \cos{x + \sin x}}{1 + \sin x}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.