Question

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

• 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 $2n\pi \pm \frac{\pi }{3}$

According to the given condition, if $a,b,c$ are in G.P., we have

$ac=b^2$

Similarly,

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

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

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

We get $\cos \theta =\frac{1}{2}$ by observation, since its value has to lie between $0$ and $1$.

Since $\cos \theta =\frac{1}{2}$, the general solution becomes $2n\pi \pm \frac{\pi }{3}$