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)

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