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: $\frac{1}{3}$
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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