# Solve this

Question:

If $\sqrt{3} \tan \theta=3 \sin \theta$ then $\left(\sin ^{2} \theta-\cos ^{2} \theta\right)=?$

(a) $\frac{1}{3}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

(d) $\frac{2}{\sqrt{3}}$

Solution:

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

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

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

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

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

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

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

$\Rightarrow \sin \theta=0$ or $1-\sqrt{3} \cos \theta=0$

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

$\Rightarrow \sin \theta=0$ or $\cos ^{2} \theta=\frac{1}{3}$

$\Rightarrow \sin \theta=0$ or $1-\cos ^{2} \theta=1-\frac{1}{3}$

$\Rightarrow \sin \theta=0$ or $\sin ^{2} \theta=\frac{2}{3}$

$\Rightarrow \sin \theta=0$ or $\sin \theta=\sqrt{\frac{2}{3}}$

For $\sin \theta=0$

$\Rightarrow \sin ^{2} \theta=0$

$\Rightarrow 1-\sin ^{2} \theta=1-0$

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

Thus, $\sin ^{2} \theta-\cos ^{2} \theta=-1$

For $\sin \theta=\sqrt{\frac{2}{3}}$,

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

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

$\Rightarrow \cos ^{2} \theta=\frac{1}{3}$

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

Hence, the correct option is (a).