Question : The expression $(\tan \theta+\cot \theta)(\sec \theta+\tan \theta)(1-\sin \theta), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $\sec \theta$
Option 2: $\operatorname{cosec} \theta$
Option 3: $\cot \theta$
Option 4: $\sin \theta$
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Correct Answer: $\operatorname{cosec} \theta$
Solution : Using the trigonometric identities $\tan \theta = \frac{\sin \theta}{\cos \theta}$, $\cot \theta = \frac{\cos \theta}{\sin \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, and $1 - \sin^2 \theta = \cos^2 \theta$. Substituting these identities into the expression, $=(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta})(\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta})(1-\sin \theta)$ Simplifying this expression, $=(\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta})(\frac{\sin \theta + 1}{\cos \theta})(1-\sin \theta)$ Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, $=(\frac{1}{\sin \theta \cos \theta})(\frac{(1-\sin ^2\theta)}{\cos \theta})$ $=(\frac{1}{\sin \theta \cos \theta})(\frac{(cos ^2\theta)}{\cos \theta})$ $=\frac{1}{\sin \theta} $ $=\operatorname{cosec} \theta$ Hence, the correct answer is $\operatorname{cosec} \theta$.
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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:
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 : 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 : 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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