If the sum of the series $\tan ^{-1} 3+\tan ^{-1} \frac{9}{2}+\tan ^{-1} \frac{33}{4}+\ldots \ldots$ to $n$ terms $=$ $\frac{(2 n+1) \pi}{A}+B \tan ^{-1}\left(2^n\right)$ then $(A+B)$ is equal to

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

If the sum of the series

\tan ^{-1} 3+\tan ^{-1} \frac{9}{2}+\tan ^{-1} \frac{33}{4}+\ldots \ldots

n

terms

=

\frac{(2 n+1) \pi}{A}+B \tan ^{-1}\left(2^n\right)

then

(A+B)

is equal to

View answer and explanation

Correct option: D

Option A (-√(5/3) < x < √(5/3)) is correct because solving the inequality leads to the condition x² < 5/3, which gives the specified range for x.

Practice All Chapter MCQs →

\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)