What is {(1+sinθ /cosθ )²+(cosθ /1+sinθ )²-2} equal to, where 0

Correct Answer is : 2tanθ2 tan θ2tanθ

Correct Answer is : 2tanθ2 tan θ2tanθ

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UPSC CDS 2 2022 Math Paper

Question : 60 of 100

Marks: +1, -0

What is

\sqrt{(\frac{1+sin\theta }{cos\theta })^{2}+(\frac{cos\theta }{1+sin\theta })^{2}-2}

\frac{\pi }{2}

Solution:

Concept:

The expression inside the square root is rearranged into a perfect square using

(a-b)^2 = a^2 + b^2 - 2ab

Explanation:

Let

a = \frac{1+\sin\theta}{\cos\theta}

and

b = \frac{\cos\theta}{1+\sin\theta}

The given expression becomes

\sqrt{a^2 + b^2 - 2ab} = \sqrt{(a-b)^2}

For

0 < \theta < \frac{\pi}{2}

a > b

, so

\sqrt{(a-b)^2} = a-b

Now

a-b = \frac{1+\sin\theta}{\cos\theta} - \frac{\cos\theta}{1+\sin\theta}

Rewrite

\frac{1+\sin\theta}{\cos\theta} = \sec\theta + \tan\theta

Using

(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1

, we have

\frac{\cos\theta}{1+\sin\theta} = \frac{1}{\sec\theta+\tan\theta} = \sec\theta - \tan\theta

Thus

a-b = (\sec\theta+\tan\theta) - (\sec\theta - \tan\theta) = 2\tan\theta

Therefore the square root simplifies to

2\tan\theta

Answer:

2\tan\theta

(Option C)

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