Question

If$\cos (x-y),\cos x$ and $\cos (x+y)$ are in HP, then $|\cos x\cdot \sec \frac{y}{2}|$equals

• A. $1$
• B. $2$
• C. $\sqrt{2}$
• D. none of these

Answer 1

Answer: (C)

$\sqrt{2}$

Since $\cos (x-y),\cos x,\cos (x+y)$ are in H.P.

We have

$\cos x=\frac{2\cos (x-y)\cos (x+y)}{\cos (x-y)+\cos (x+y)}=\frac{2({\cos }^{2}x-{\sin }^{2}y)}{2\cos x\cos y}$

$\Rightarrow {\cos }^{2}x\cos y={\cos }^{2}x-{\sin }^{2}y$

$\Rightarrow (1-\cos y){\cos }^{2}x={\sin }^{2}y$

$\Rightarrow {\cos }^{2}x=\frac{{\sin }^{2}y}{1-\cos y}=\frac{4{\sin }^2\left(\frac{y}{2}\right){\cos }^2\left(\frac{y}{2}\right)}{2{\sin }^2\left(\frac{y}{2}\right)}$

$\Rightarrow {\cos }^{2}x=2{\cos }^2\left(\frac{y}{2}\right)\Rightarrow \cos x\sec \left(\frac{y}{2}\right)=\pm \sqrt{2}$