JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

The value of

\cot \left(\sum_{n=1}^{19} \cot ^{-1}\left(1+\sum_{p=1}^{n} 2 p\right)\right)

is:

View answer and explanation

Correct option: A

\frac{21}{19}

Option A (

\frac{{21}}{{19}}

) is correct because by simplifying the summation and using the arctangent subtraction formula, the expression reduces to

\frac{{21}}{{19}}

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

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

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\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)