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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Hard

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Analyzing +4 | -1

The value of

\sum \limits_{r = 0}^\infty {\tan ^{ - 1}}\left( {\frac{1}{{1 + r + {r^2}}}} \right)\;

is equal to

View answer and explanation

Correct option: A

\frac{\pi }{2}

Option A (

\frac{\pi }{2}

) is correct because the sum telescopes to

\tan^{-1}(\infty)

, which equals

\frac{\pi }{2}

.

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

is- The value of

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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},

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