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Expression 1+sin⁡θ1−sin⁡θ \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}1−sinθ1+sinθ​​​ is equal to:
Question

Expression 1+sin⁡θ1−sin⁡θ \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}​ is equal to:

A.

secθ - cot θ

B.

secθ - tan⁡θ

C.

sec θ + tan θ

D.

cosec θ + cotθ

Correct option is C

Given:
Expression: 1+sin⁡θ1−sin⁡θ \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}​​
Formula Used:
​sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1​​

​sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}​​

​tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}​​
Solution:
Multiply the numerator and the denominator inside the square root by (1+sin⁡θ):(1 + \sin\theta):​​
​=(1+sin⁡θ)(1+sin⁡θ)(1−sin⁡θ)(1+sin⁡θ)= \sqrt{\frac{(1+\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}}​​

​=(1+sin⁡θ)21−sin⁡2θ= \sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}​​

Since 1−sin⁡2θ=cos⁡2θ:1-\sin^2\theta = \cos^2\theta:​​

​=(1+sin⁡θ)2cos⁡2θ= \sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}​​
Taking the square root:
​=1+sin⁡θcos⁡θ= \frac{1+\sin\theta}{\cos\theta}​​
Separate the terms:
​=1cos⁡θ+sin⁡θcos⁡θ= \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}​​
​=sec⁡θ+tan⁡θ= \sec\theta + \tan\theta​​
Final Answer
So the correct answer is (c)

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