Question : If $\sqrt{3}\tan\theta=3\sin\theta$, then the value of $(\sin^{2}\theta-\cos^{2}\theta)$ is:
Option 1: $1$
Option 2: $3$
Option 3: $\frac{1}{3}$
Option 4: None
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Correct Answer: $\frac{1}{3}$
Solution : Given: $\sqrt{3}\tan\theta=3\sin\theta$ ⇒ $\sqrt{3}\frac{\sin\theta}{\cos\theta}=3\sin\theta$ ⇒ $\cos \theta=\frac{1}{\sqrt{3}}$ So, $\sin^{2}\theta-\cos^{2}\theta=1-\cos^{2} \theta-\cos^{2} \theta=1-2\cos^{2} \theta$ Putting the value of $\cos \theta=\frac{1}{\sqrt{3}}$, we have, = $1-2×(\frac{1}{\sqrt{3}})^{2}$ = $1-\frac{2}{3}$ = $\frac{1}{3}$ Hence, the correct answer is $\frac{1}{3}$.
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