Q. If $\frac{\sin\theta}{6},\cos{\theta\ and\tan{\theta\ }}$are in GP, then the general value of $\theta$ is
A) $2n\pi \pm \frac{\pi}{3},\ n \in I$
B) $2n\pi \pm \frac{\pi}{6},\ n \in I$
C) $2n\pi + ( - 1)^{n}\ \frac{\pi}{3},\ n \in I$
D) $n\ \pi + \frac{\pi}{3},n \in I$
Explanation
(a)
Since, $\cos^{2}{\theta = \frac{1}{6}\sin{\theta \bullet \tan\theta}}$
$\Rightarrow 6\cos^{3}{\theta + \cos^{2}{\theta - 1 = 0}}$
As $\cos{\theta = \frac{1}{2}}$ satisfied the equation.
$\therefore(2\cos{\theta - 1)(3\cos^{2}{\theta + 2\cos{\theta + 1) = 0}}}$
$\Rightarrow \cos{\theta = \frac{1}{2}}$ (other values of $\cos\theta$ are imaginary)
$\therefore\ \theta = 2n\pi \pm \frac{\pi}{3},\ n \in I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.