JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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\sin ^{-1} \frac{1}{\sqrt{2}}+\sin ^{-1} \frac{\sqrt{2}-1}{\sqrt{6}}+\sin ^{-1} \frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\sin ^{-1} \frac{2-\sqrt{3}}{\sqrt{20}}+\sin ^{-1} \frac{\sqrt{5}-2}{\sqrt{30}}+\ldots \infty

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Correct option: B

\frac{\pi}{2}

The correct answer is

\frac{\pi}{2}

because the given series telescopes to

\frac{\pi}{2}

n

approaches infinity.

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x

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2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)