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