Question

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

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

Answer 1

The correct option is A

$\pm \sqrt{2}$

Explanation for the correct option:

Step 1: Form the equation using common difference of AP

Given, $\sec (x-y),\sec x,\sec (x+y)$ are in AP.
$\Rightarrow 2\sec x=\sec (x-y)+\sec (x+y)$
$$\begin{gathered} \Rightarrow \frac{2}{\cos x}=\frac{1}{\cos (x-y)}+\frac{1}{\cos (x+y)} \\ \Rightarrow 2\cos (x-y)\cos (x+y)=\left[\cos \right(x+y)+\cos (x-y\left)\right]\cos x \\ \Rightarrow \cos \left(2x\right)+\cos \left(2y\right)=\left(2\cos x\cos y\right)\cos x \\ \Rightarrow 2{\cos }^{2}x-2+2{\cos }^{2}y=2{\cos }^{2}x\cos y \\ \Rightarrow {\cos }^{2}x+{\cos }^{2}y-1={\cos }^{2}x\cos y \\ \Rightarrow {\cos }^{2}x-{\cos }^{2}x\cos y+\left({\cos }^{2}y-1\right)=0 \\ \Rightarrow {\cos }^{2}x(1-\cos y)-1(1-\cos y)(1+\cos y)=0 \\ \Rightarrow (1-\cos y)\left({\cos }^{2}x-1-\cos y\right)=0 \end{gathered}$$

Step 2: Find the value of the expression $\cos x\sec \left(\frac{y}{2}\right)$

By using the zero property of product we have:
$\cos y=1\mathrm{or}{\cos }^{2}x-1-\cos y=0$
$\Rightarrow y=0or\cos x=\pm \sqrt{2}$

Therefore, the value of $\cos x\sec \left(\frac{y}{2}\right)$ is:
$\begin{array}{rcl}\cos x\sec \left(\frac{y}{2}\right)& =& \pm \sqrt{2}\sec 0°\\ & =& \pm \sqrt{2}\end{array}$
Hence, option A is correct.