Question : If $3\left(\cot ^2 \theta-\cos ^2 \theta\right)=1-\sin ^2 \theta, 0^{\circ}<\theta<90^{\circ}$, then $\theta$ is equal to:
Option 1: $30^{\circ}$
Option 2: $60^{\circ}$
Option 3: $45^{\circ}$
Option 4: $15^{\circ}$
Correct Answer: $60^{\circ}$
Solution :
Given: $3\left(\cot ^2 \theta-\cos ^2 \theta\right)=1-\sin ^2 \theta$
⇒ $3\left(\cot ^2 \theta-\cos ^2 \theta\right)=\cos ^2 \theta$
⇒ $3\cot ^2 \theta = 4\cos ^2 \theta$
⇒ $\frac{3\cos^2\theta}{\sin^2 \theta} = 4\cos ^2 \theta$
⇒ $\sin^2 \theta = \frac{3}{4}$
⇒ $\sin \theta = \frac{\sqrt3}{2}$
⇒ $\theta = 60^{\circ}$
Hence, the correct answer is $60^{\circ}$.
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