Question : The expression $\frac{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+\cot \theta+1)(\operatorname{cosec} \theta-\cot \theta+1)-2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $\frac{1}{2} \sin \theta$
Option 2: $2 \sin \theta$
Option 3: $\sec \theta$
Option 4: $2\operatorname{cosec} \theta$
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Correct Answer: $2 \sin \theta$
Solution :
$\frac{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+\cot \theta+1)(\operatorname{cosec} \theta-\cot \theta+1)-2}, 0^{\circ}<\theta<90^{\circ}$
$=\frac{(cos^2\theta)^2-(\sin^2\theta)^2+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+1+\cot \theta)(\operatorname{cosec} \theta+1-\cot \theta)-2}$
$=\frac{(cos^2+\sin^2\theta)(cos^2-\sin^2\theta)+2 \sin ^2 \theta+3}{[(\operatorname{cosec} \theta+1)^2-\cot^2 \theta]-2}$
$=\frac{cos^2+\sin^2\theta +3}{[\operatorname{cosec^2} \theta+1+2\operatorname{cosec \theta}-\cot^2 \theta]-2}$
$=\frac{4}{2+2\operatorname{cosec} \theta-2}$
$=\frac{4}{2\operatorname{cosec} \theta}$
$=2 \sin \theta$
Hence, the correct answer is $2 \sin \theta$.
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