Question

If $\cos (x-y)$, $\cos x$ and $\cos (x+y)$ are in H.P, then value of $\cos x\sec \left(\frac{y}{2}\right)$ is

• A. $\pm \sqrt{2}$
• B. $\pm \sqrt{3}$
• C. $\pm 2$
• D. $\pm 1$

Answer 1

Answer: (A)

$\pm \sqrt{2}$

$cos(x-y),cos\left(x\right),cos(x+y)$ are in H.P
$\Rightarrow cos\left(x\right)=\frac{2cos(x-y)cos(x+y)}{\left(cos\right(x-y)+cos(x+y\left)\right)}$ (from the definition of Harmonic mean)
$\Rightarrow cos\left(x\right)=\frac{\left(cos\right(2x)+cos(2y\left)\right)}{\left(2cos\right(x\left)cos\right(y\left)\right)}$
$\Rightarrow 2co{s}^2\left(x\right)cos\left(y\right)=2co{s}^2\left(x\right)-1+2co{s}^2\left(y\right)-1$
$\Rightarrow 2co{s}^2\left(x\right)\left(cos\right(y)-1)=2\left(2co{s}^2\right(y)-1)$
$\Rightarrow co{s}^2\left(x\right)\left(cos\right(y)-1)=\left(cos\right(y)+1)\left(cos\right(y)-1)$
$\Rightarrow co{s}^2\left(x\right)=cos\left(y\right)+1$
$\Rightarrow co{s}^2\left(x\right)=2co{s}^2\left(y/2\right)$
$\Rightarrow co{s}^2\left(x\right)se{c}^2\left(y/2\right)=2$
$\therefore cos\left(x\right)sec\left(y/2\right)=\sqrt{2}or-\sqrt{2}$