Question : Find $\frac{1}{ \sin \theta}-\sin \theta$.
Option 1: $\cos \theta \cot \theta$
Option 2: $\cos \theta \sec \theta$
Option 3: $\cos \theta \operatorname{cosec} \theta$
Option 4: $\cos \theta \tan \theta$
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Correct Answer: $\cos \theta \cot \theta$
Solution : Given: $\frac{1}{ \sin \theta}-\sin \theta$ $= \frac{1-\sin^2\theta}{\sin\theta}$ $= \frac{\cos^2\theta}{\sin\theta}$ $= \cot\theta \cos\theta$ Hence, the correct answer is $\cot\theta \cos\theta$.
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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 : Which of the following is equal to $[\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}]$?
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 : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:
Question : The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is:
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