Q. If $\sec{\theta + \tan{\theta = k,\cos\theta}}$ equals to
A) $\frac{k^{2} + 1}{2k}$
B) $\frac{2k}{k^{2} + 1}$
C) $\frac{k}{k^{2} + 1}$
D) $\frac{k}{k^{2} - 1}$
Explanation
(b)
$\sec{\theta + \tan{\theta = k}}$ …(i)
$\Rightarrow \ \ \frac{\sec^{2}{\theta - \tan^{2}\theta}}{\sec{\theta - \tan\theta}} = k$
$\Rightarrow \sec{\theta - \tan{\theta = \frac{1}{k}}}$ …(ii)
On adding Eqs. (i) and (ii), we get
$2\sec{\theta = k + \frac{1}{k} = \frac{k^{2} + 1}{k}}$
$\Rightarrow \cos{\theta = \frac{2k}{k^{2} + 1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.