Question

If $\sin (x-y), \sin x$ and $\sin (x+y)$ are in H.P., then $\sin x \cdot \sec \frac{y}{2}=$

• A. 2
• B. $\sqrt{2}$
• C. $-\sqrt{2}$
• D. -2

Answer 1

Answer: (B), (C)

$ \sin x=\frac{2 \sin (x-y) \sin (x+y)}{\sin (x-y)+\sin (x+y)}=\frac{2\left(\sin ^2 x-\sin ^2 y\right)}{2 \sin x \cos y}$
$\sin ^2 x \cos y=\sin ^2 x-\sin ^2 y \sin ^2 x(1-\cos y)=\sin ^2 y$
$\sin ^2 x \cdot 2 \sin ^2 \frac{y}{2}=4 \sin ^2 \frac{y}{2} \cos ^2 \frac{y}{2} ; \sin ^2 x \sec ^2 \frac{y}{2}=2 \sin x \sec \frac{y}{2}= \pm \sqrt{2}$