Question

If $\frac{1}{6}\sin \theta ,\cos \theta$ and $\tan \theta$ are in geometric progression, then the solution set of $\theta$ is

• A. 2$n\pi \pm \left(\frac{\pi }{6}\right)$
• B. 2$n\pi \pm \left(\frac{\pi }{3}\right)$
• C. $n\pi +(-1{)}^n\left(\frac{\pi }{3}\right)$
• D. $n\pi +\left(\frac{\pi }{3}\right)$

Answer 1

Answer: (B)

2$n\pi \pm \left(\frac{\pi }{3}\right)$

$\frac{1}{6}\sin \theta ,\cos \theta$ and $\tan \theta$ are in G.P.

$\therefore$ ${\cos }^2\theta =\frac{1}{6}\sin \theta \tan \theta$

$\Rightarrow$ ${\cos }^2\theta =\frac{1}{6}\frac{{\sin }^2\theta }{\cos \theta }$

$\Rightarrow$ ${\cos }^3\theta =\frac{{\sin }^2\theta }{6}$

$\Rightarrow$ $6{\cos }^3\theta =(1-{\cos }^2\theta )$

$\Rightarrow$ $6{\cos }^3\theta +{\cos }^2\theta -1=0$

$\Rightarrow$ $(\cos \theta -\frac{1}{2})(6{\cos }^2\theta +4\cos \theta +2)=0$

$\Rightarrow$ $2(\cos \theta -\frac{1}{2})(3{\cos }^2\theta +2\cos \theta +1)=0$

$\Rightarrow$ $\cos \theta -\frac{1}{2}=0$

$\Rightarrow$ $\cos \theta =\frac{1}{2}$

$\Rightarrow$ $\cos \theta =\cos \frac{\pi }{3}$

$\Rightarrow$ $\theta =2n\pi \pm \frac{\pi }{3}$