Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Option 1: 0
Option 2: 2
Option 3: 5
Option 4: 1
Correct Answer: 1
Solution :
$\cot \theta=\frac{1}{\sqrt{3}}$
⇒ $\theta=60^{\circ}$
So, $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$
= $\frac{2-\sin ^2 60^{\circ}}{1-\cos ^2 60^{\circ}}+\left(\operatorname{cosec}^2 60^{\circ} -\sec 60^{\circ}\right)$
= $\frac{2- \frac{3}{4}}{1-\frac{1}{4}}+\frac{4}{3} -2$
= $\frac{\frac{5}{4}}{\frac{3}{4}}+\frac{4}{3} -2$
= $\frac{5}{3}+\frac{4}{3} -2$
= $\frac{9}{3} -2$
= 3 – 2
= 1
Hence, the correct answer is 1.
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