MCQ
Evaluate $: \sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}$
- A$2\sin\theta$
- B$2 \cos \theta$
- ✓2 cosec $\theta$
- D$2 \sec \theta$
✓
Answer
Correct option: C.
2 cosec $\theta$
(c) : We have, $\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}$
$
=\frac{(\sec \theta-1)+(\sec \theta+1)}{\sqrt{\sec ^2 \theta-1}}=\frac{2 \sec \theta}{\sqrt{\tan ^2 \theta}} \quad\left[\because \sec ^2 \theta-1=\tan ^2 \theta\right]
$
$
=\frac{2 \sec \theta}{\tan \theta}=\frac{2}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}=\frac{2}{\sin \theta}=2 \operatorname{cosec} \theta
$
$
=\frac{(\sec \theta-1)+(\sec \theta+1)}{\sqrt{\sec ^2 \theta-1}}=\frac{2 \sec \theta}{\sqrt{\tan ^2 \theta}} \quad\left[\because \sec ^2 \theta-1=\tan ^2 \theta\right]
$
$
=\frac{2 \sec \theta}{\tan \theta}=\frac{2}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}=\frac{2}{\sin \theta}=2 \operatorname{cosec} \theta
$
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