JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

S=\tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots

+\tan ^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)

, then

\tan S

is equal to:

View answer and explanation

Correct option: C

\frac{20}{n^2+20 n+1}

Option C (

\frac{20}{n^2+20n+1}

) is correct because the sum telescopes to

\tan^{-1}(\frac{n+20}{n+1}) - \tan^{-1}(\frac{n}{n+1})

, and applying the tangent function simplifies to the given expression.

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)