Question

If $\cos (x-y),\cos x$ and $\cos (x+y)$ are in $H.P,$ then $\cos x\sec \left(y/2\right)=$............

• A. $\pm \sqrt{3}$
• B. $\sqrt{2}$
• C. $-\sqrt{3}$
• D. $\pm \sqrt{2}$

Answer 1

The correct option is D $\pm \sqrt{2}$

Since $\cos (x-y),\cos x,\cos (x+y)$ are in H.P
We have
$\cos x=\frac{2\cos (x-y)\cos (x+y)}{\cos (x-y)+\cos (x+y)}=\frac{2({\cos }^{2}x-{\sin }^{2}y)}{2\cos x\cos y}$
$\Rightarrow {\cos }^{2}x\cos y={\cos }^{2}x-{\sin }^{2}y\Rightarrow (1-\cos y){\cos }^{2}x={\sin }^{2}y$
$\Rightarrow {\cos }^{2}x=\frac{{\sin }^{2}y}{1-\cos y}=\frac{4{\sin }^2\left(\frac{y}{2}\right){\cos }^2\left(\frac{y}{2}\right)}{2{\sin }^2\left(\frac{y}{2}\right)}$
$\Rightarrow {\cos }^{2}x=2{\cos }^2\left(\frac{y}{2}\right)\Rightarrow \cos x\sec \left(\frac{y}{2}\right)=\pm \sqrt{2}$