JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Analyzing +4 | -1

The sum of the infinite series

{\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 2 - 1}}{{\sqrt 6 }}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt {12} }}} \right) + ...

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

View answer and explanation

Correct option: C

\frac{\pi }{2}

Option C (

\frac{\pi}{2}

) is correct because the series telescopes to

{\tan^{-1}}(\sqrt{n})

which approaches

\frac{\pi}{2}

as n approaches infinity.

More Questions from INVERSE TRIGONOMETRY

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\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

\sin \left(2 \tan ^{-1}(1 / 3)\right)+

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

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)