Question

Find the general value of $\theta$ which is satisfies both $\sin \theta =\frac{-1}{2}$ and $\tan \theta =\frac{1}{\sqrt{3}}$ simultaneously.

• A. $\theta =\frac{4\pi }{3}$
• B. $\theta =\frac{\pi }{3}$
• C. $\theta =-\frac{4\pi }{3}$
• D. None of these

Answer 1

The correct option is A $\theta =\frac{4\pi }{3}$

Consider the given value ,

$\sin \theta =\frac{-1}{2}$

$\sin \theta =-\sin \frac{\pi }{6}=\sin (\pi +\frac{\pi }{6})$

$\sin \theta =\sin (\pi +\frac{\pi }{6})=\sin \frac{7\pi }{6}$

$\sin \theta =\sin \frac{7\pi }{6}$

$\theta =\frac{7\pi }{6}$

or$\sin \theta =\sin (-\frac{\pi }{6})$

$\theta =-\frac{\pi }{6}$

Again consider,

$\tan \theta =\sqrt{3}$

$\tan \theta =\tan \frac{\pi }{3}$

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

or $\tan \theta =\sqrt{3}$

$\tan \theta =\tan \frac{\pi }{3}=\tan (\pi +\frac{\pi }{3})$

$\tan \theta =\tan (\pi +\frac{\pi }{3})=\tan \frac{4\pi }{3}$

$\tan \theta =\tan \frac{4\pi }{3}$

$\theta =\frac{4\pi }{3}$

Hence , there is no any general value of $\theta$ for which given equation will become satisfied