Question

Suppose $\theta$ and $\phi\left(\ne\,0\right)$ are such that sec$\left(\theta+\phi\right), sec\,\theta$ and $sec\left(\theta-\phi\right)$ are in A.P. If $cos\,\theta=k\,cos\left(\frac{\phi}{2}\right)$ for some $k$, then $k$ is equal to

• A. $\pm\sqrt{2}$
• B. $\pm1$
• C. $\pm\frac{1}{\sqrt{2}}$
• D. $\pm2$

Answer 1

Answer: (A)

Since, $sec\left(\theta-\phi\right), sec\theta$ and $ec\left(\theta+\phi\right)$ are in $A.P.,$
$\therefore 2\,sec\theta=sec\left(\theta-\phi\right)+sec\left(\theta +\phi \right)$
$\Rightarrow \frac{1}{cos\,\theta}=\frac{cos\left(\theta+\phi \right)+cos\left(\theta +\phi \right)}{cos\left(\theta -\phi \right)cos\left(\theta +\phi \right)}$
$\Rightarrow 2\left(cos^{2}\,\theta-sin^{2}\,\phi\right)=cos\,\theta\left[2\,cos\,\theta\,cos\,\phi\quad\right]$
$\Rightarrow cos^{2}\,\theta=1+cos\,\phi=2\,cos^{2} \frac{\phi}{2}$
$\therefore cos\,\theta=\pm\sqrt{2}cos \frac{\phi}{2}$
But given $cos\theta=k\,cos \frac{\phi}{2}$
$\therefore k=\pm\sqrt{2}$