Question

If $\frac{1}{6}\sin \theta ,\cos \theta ,\tan \theta$ are in G.P., then $\theta$ is equal to $\left(nϵz\right)$

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

Answer 1

The correct option is A $2n\pi \pm \frac{\pi }{3}$

$\frac{1}{6}sin\theta tan\theta =co{s}^2\theta$

$sin\theta tan\theta =6co{s}^2\theta$

$sin\theta \times \frac{sin\theta }{cos\theta }=6co{s}^2\theta$

$si{n}^2\theta =6co{s}^3\theta$

$1-co{s}^2\theta =6co{s}^3\theta$

$6co{s}^3\theta +co{s}^2\theta -1=0$

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

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