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:
Option 1: $\sin \theta$
Option 2: $2 \cos \theta$
Option 3: $\cot \theta$
Option 4: $2 \tan \theta$
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Correct Answer: $2 \tan \theta$
Solution : $\frac{(1-\sin \theta+\cos \theta)^2(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\sec \theta-\tan \theta)(\tan \theta+\cot \theta)}$ $=\frac{(1+\sin^2 \theta+\cos^2 \theta-2\sin\theta-2\sin\theta\cos\theta+2\cos\theta)(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta})(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta})}$ $=\frac{(2-2\sin\theta-2\sin\theta\cos\theta+2\cos\theta)(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\frac{1-\sin\theta}{\cos\theta})(\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta})}$ $=\frac{(2(1-\sin\theta)+ 2\cos\theta(1-\sin\theta)(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\frac{1-\sin\theta}{\cos\theta})(\frac{1}{\sin\theta\cos\theta})}$ $=\frac{((1-\sin\theta)2 (1+\cos\theta)(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\frac{1-\sin\theta}{\cos\theta})(\frac{1}{\sin\theta\cos\theta})}$ $=2(1-\cos^2\theta).\frac{1}{\cos\theta.\sin\theta}$ $=2\sin^2\theta.\frac{1}{\cos\theta.\sin\theta}$ $=2 \tan \theta$ Hence, the correct answer is $2 \tan \theta$.
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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:
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Question : The expression $\frac{\tan ^6 \theta-\sec ^6 \theta+3 \sec ^2 \theta \tan ^2 \theta}{\tan ^2 \theta+\cot ^2 \theta+2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $(\tan \theta+\cot \theta)(\sec \theta+\tan \theta)(1-\sin \theta), 0^{\circ}<\theta<90^{\circ}$, is equal to:
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