Question : The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is:
Option 1: $\sec\theta+\tan \theta$
Option 2: $\operatorname{cosec} \theta-\cot \theta$
Option 3: $\operatorname{cosec} \theta+\cot \theta$
Option 4: $\sec\theta-\tan \theta$
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Correct Answer: $\operatorname{cosec} \theta+\cot \theta$
Solution : Given: $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ = $\sqrt{\frac{(1+\cos \theta)(1+\cos \theta)}{(1-\cos \theta)(1+\cos \theta)}}$ = $\sqrt{\frac{(1+\cos \theta)^2}{1-\cos^2\theta}}$ = $\sqrt{\frac{(1+\cos \theta)^2}{\sin^2\theta}}$ = $\frac{(1+\cos \theta)}{\sin\theta}$ = $\operatorname{cosec} \theta+\cot \theta$ Hence, the correct answer is $\operatorname{cosec}\theta+\cot\theta$.
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Question : For any real values of $\theta, \sqrt{\frac{\sec\theta \:-\: 1}{\sec\theta \:+\: 1}}=$?
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 : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ 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:
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