Question : If $\sqrt{3} \tan \theta=3 \sin \theta$, then what is the value of $\sin ^2 \theta-\cos ^2 \theta$?
Option 1: $\frac{1}{5}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{1}{2}$
Option 4: $\frac{1}{3}$
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Correct Answer: $\frac{1}{3}$
Solution : Given: $\sqrt{3} \tan \theta=3 \sin \theta$ ⇒ $\sqrt{3}\times \frac{\sin \theta}{\cos\theta}=3 \sin \theta$ ⇒ $\frac{\sqrt{3}}{3}=\cos\theta$ ⇒ $\cos\theta=\frac{1}{\sqrt3}$ Squaring both sides of the above equation, $\cos^2 \theta=\frac{1}{3}$ (equation 1) We know the trigonometric identity, $\sin ^2 \theta+\cos ^2 \theta=1$. Substitute the value from the equation 1 in the above identity, ⇒ $\sin ^2 \theta+\frac{1}{3}=1$ ⇒ $\sin ^2 \theta=1-\frac{1}{3}$ ⇒ $\sin ^2 \theta=\frac{2}{3}$ The value of $\sin ^2 \theta-\cos ^2 \theta=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$. Hence, the correct answer is $\frac{1}{3}$.
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