NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

The value of $\sin^{-1} \frac{1}{\sqrt{2}} + \sin^{-1} \frac{\sqrt{2}-\sqrt{1}}{\sqrt{6}} + \sin^{-1} \frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}} + \cdots$ to infinity is equal to:

Choose the correct answer:

Explanation

1. Analyze the General Term ( $T_n$ ):

The $n^{th}$ term of the series is:

T_n = \sin^{-1} \left( \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}} \right)

2. Simplify using Identity:

Using the identity $\sin^{-1} x - \sin^{-1} y = \sin^{-1} (x\sqrt{1-y^2} - y\sqrt{1-x^2})$ , we can rewrite the terms.

Alternatively, observe that:

\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)

Let's convert the sine terms to tangent terms for easier telescoping:

T_1 = \sin^{-1} \frac{1}{\sqrt{2}} = \tan^{-1} (1)

T_2 = \sin^{-1} \frac{\sqrt{2}-1}{\sqrt{6}} = \tan^{-1} \left( \frac{1}{\sqrt{1}} \right) - \tan^{-1} \left( \frac{1}{\sqrt{2}} \right) \text{ (Incorrect approach, let's use the standard property)}

Using the property: $\sin^{-1} \left( \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}} \right) = \tan^{-1} \sqrt{n} - \tan^{-1} \sqrt{n-1}$

3. Expansion of the Series:

S = \sum_{n=1}^{\infty} (\tan^{-1} \sqrt{n} - \tan^{-1} \sqrt{n-1})

S = (\tan^{-1} \sqrt{1} - \tan^{-1} \sqrt{0}) + (\tan^{-1} \sqrt{2} - \tan^{-1} \sqrt{1}) + (\tan^{-1} \sqrt{3} - \tan^{-1} \sqrt{2}) + \cdots + (\tan^{-1} \sqrt{\infty} - \cdots)

4. Telescoping Sum:

Notice that all intermediate terms cancel out:

S = -\tan^{-1} (0) + \tan^{-1} (\infty)

S = 0 + \frac{\pi}{2}

S = \frac{\pi}{2}

Conclusion:

The value of the infinite series is $\frac{\pi}{2}$ . The correct option is (c).