JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Sum to

n

terms of the series

\operatorname{cosec}^{-1} \sqrt{10}+\operatorname{cosec}^{-1} \sqrt{50}+\operatorname{cosec}^{-1} \sqrt{170}+\ldots \ldots \\ +\operatorname{cosec}^{-1} \sqrt{\left(n^2+1\right)\left(n^2+2 n+2\right)} \text { is }

View answer and explanation

Correct option: A

\tan ^{-1}(n+1)-\frac{\pi}{4}

Option A is correct because the sum of the series can be simplified using the arctangent subtraction formula, resulting in

\tan^{-1}(n+1) - \frac{\pi}{4}

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x

{x^2} + {y^2} + {z^2} = {r^2},

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

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