Question

If $\sec (x-y),\sec x,\sec (x+y)$ are in A.P. and $\sec y\ne 1$, then the angle $y$ can lie in the

• A. $I$ quadrant
• B. $II$ quadrant
• C. $III$ quadrant
• D. $IV$ quadrant

Answer 1

Answer: (B), (C)

The correct options are
B $II$ quadrant
C $III$ quadrant

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