Question

If $\sec (x-y),\sec x,\sec (x+y)$ are in A.P., then $\cos x\sec \left(\frac{y}{2}\right)=$

• A. $\pm 2$
• B. $\pm \frac{1}{\sqrt{2}}$
• C. $\pm \frac{1}{2}$
• D. $\pm \sqrt{2}$

Answer 1

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

Since $\sec (x-y),\sec x,\sec (x+y)$ are in AP

$2\sec x=\sec (x-y)+\sec (x+y)\frac{2}{\cos x}=\frac{1}{\cos (x-y)}+\frac{1}{\cos (x+y)}2\cos (x-y)\cos (x+y)=[\cos (x+y)+\cos (x-y)]\cos x\cos \left(2x\right)+\cos \left(2y\right)=\left(2\cos x\cos y\right)\cos x2{\cos }^{2}x-2+2{\cos }^{2}y=2{\cos }^{2}x\cos y{\cos }^{2}x+{\cos }^{2}y-1={\cos }^{2}x\cos y{\cos }^{2}x-{\cos }^{2}x\cos y+({\cos }^{2}y-1)=0{\cos }^{2}x(1-\cos y)-1(1-\cos y)(1+\cos y)=0(1-\cos y)({\cos }^{2}x-1-\cos y)=0$

which gives $\cos y=1y=0$

${\cos }^{2}x-1-\cos y=0{\cos }^{2}x=1+\cos y=1+1\cos x=\pm \sqrt{2}$

then, $\cos x\sec \left(\frac{y}{2}\right)=\pm \sqrt{2}\sec 0^o=\pm \sqrt{2}$