Question

If $\sec (x-y),\sec x,\sec (x+y)$ are in arithmetic progression and $\sec y\ne 1,$ then the angle $y$ can be

• A. $\frac{\pi }{5}$
• B. $\frac{7\pi }{12}$
• C. $\frac{13\pi }{7}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B $\frac{7\pi }{12}$

$\sec (x-y),\sec x,\sec (x+y)$ are in A.P.
$\Rightarrow 2\sec x=\sec (x+y)+\sec (x-y)$
$\Rightarrow \cos x=\frac{2\cos (x-y)\cos (x+y)}{\cos (x-y)+\cos (x+y)}$
$\Rightarrow \cos x=\frac{\cos 2x+\cos 2y}{2\cos x\cos y}$
$\Rightarrow 2{\cos }^{2}x\cos y=\cos 2x+\cos 2y$
$\Rightarrow 2{\cos }^{2}x\cos y=(2{\cos }^{2}x-1)+(2{\cos }^{2}y-1)\Rightarrow 2{\cos }^{2}x(\cos y-1)=2({\cos }^{2}y-1)$
As $\sec y\ne 1,$
${\cos }^{2}x=\cos y+1$
$\Rightarrow \cos y=-{\sin }^{2}x$
So, $\cos y<0$
Therefore, $y$ can lie either in II or III quadrant.