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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Question : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $\frac{(1-\sin \theta+\cos \theta)^2(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\sec \theta-\tan \theta)(\tan \theta+\cot \theta)}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : Let $0^{\circ}<\theta<90^{\circ}$, $\left(1+\cot ^2 \theta\right)\left(1+\tan ^2 \theta\right) × (\sin \theta-\operatorname{cosec} \theta)(\cos \theta-\sec \theta)$ is equal to:
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