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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