Question

If $\cos (\theta -\alpha )$, $\cos \theta$ and $\cos (\theta +\alpha )$ are in$HP$, then $\cos \theta sec\frac{\alpha }{2}$ equal to

• A. $\pm \sqrt{2}$
• B. $\pm \sqrt{3}$
• C. $\pm \frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

The correct option is A

$\pm \sqrt{2}$

Explanation for the correct option:

Find the value of $\cos \theta sec\frac{\alpha }{2}$:

Given, $\cos (\theta -\alpha )$, $\cos \theta$ and $\cos (\theta +\alpha )$ are in $HP$.

Here, $a=\cos (\theta –\alpha ),b=\cos \theta$ and $c=\cos (\theta +\alpha )$

$\begin{array}{rcl}\therefore \cos \theta & =& \frac{2\cos \left(\theta -\alpha \right)\cos \left(\theta +\alpha \right)}{\cos \left(\theta -\alpha \right)+\cos \left(\theta +\alpha \right)}\end{array}$ $\left[\mathbf{\because }\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{a}\mathbf{c}}{\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}}\right]$

$\begin{array}{rcl}& =& \frac{2\left({\cos }^2\theta {\cos }^2\alpha -{\sin }^2\theta {\sin }^2\alpha \right)}{2\cos \theta \cos \alpha }\\ & =& \frac{{\cos }^2\theta {\cos }^2\alpha -{\sin }^2\theta {\sin }^2\alpha }{\cos \theta \cos \alpha }\\ & =& \frac{{\cos }^2\theta {\cos }^2\alpha -\left(1-{\cos }^2\theta \right){\sin }^2\alpha }{\cos \theta \cos \alpha }\\ & =& \frac{{\cos }^2\theta {\cos }^2\alpha -{\sin }^2\alpha +{\cos }^2\theta {\sin }^2\alpha }{\cos \theta \cos \alpha }\end{array}$

$\Rightarrow$ $\cos \theta =\frac{{\cos }^2\theta \left({\cos }^2\alpha +{\sin }^2\alpha \right)-{\sin }^2\alpha }{\cos \theta \cos \alpha }$

$\Rightarrow$ ${\cos }^2\theta \cos \alpha ={\cos }^2\theta -{\sin }^2\alpha$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\Rightarrow$ ${\sin }^2\alpha ={\cos }^2\theta \left(1-\cos \alpha \right)$

$\Rightarrow$${\left[2\sin \left(\frac{\alpha }{2}\right)\cos \left(\frac{\alpha }{2}\right)\right]}^2={\cos }^2\theta 2\left[{\sin }^2\left(\frac{\alpha }{2}\right)\right]$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathbf{sin}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{cos}\frac{\mathbf{\theta }}{\mathbf{2}}\right]$

$\Rightarrow$$4{\sin }^2\left(\frac{\alpha }{2}\right){\cos }^2\left(\frac{\alpha }{2}\right)=2{\cos }^2\theta {\sin }^2\left(\frac{\alpha }{2}\right)$

$\Rightarrow$ $2{\cos }^2\left(\frac{\alpha }{2}\right)={\cos }^2\theta$

$\Rightarrow$ $\frac{2}{se{c}^2\left(\frac{\alpha }{2}\right)}={\cos }^2\theta$

$\Rightarrow$ ${\cos }^2\theta se{c}^2\left(\frac{\alpha }{2}\right)=2$

$\mathbf{\therefore }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{}\mathit{s}\mathit{e}\mathit{c}\mathbf{}\left(\frac{\alpha }{2}\right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{\pm }\sqrt{\mathbf{2}}$

Hence, Option ‘A’ is Correct.