Question

If $\sqrt{3}\tan \theta =3\sin \theta$ then the value of ${\sin }^2\theta -{\cos }^2\theta$ is

• A. 2
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. 3

Answer 1

The correct option is C $\frac{1}{3}$

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

$\sqrt{3}\frac{\sin \theta }{\cos \theta }=3\sin \theta$

$\sqrt{3}\frac{\sin \theta }{\cos \theta }-3\sin \theta =0$

$\sqrt{3}\sin \theta (\frac{1}{\cos \theta }-\sqrt{3})=0$

Now either $\sin \theta =0$ or $\cos \theta =\frac{1}{\sqrt{3}}$

But $\sin \theta \ne 0$

Hence $\cos \theta =\frac{1}{\sqrt{3}}$

Now u\sin g ${\sin }^2\theta +{\cos }^2\theta =1$,we will get $\sin \theta =\sqrt{\frac{2}{3}}$

Hence ${\sin }^2\theta -{\cos }^2\theta =\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$