JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

0 < a,\,\,b < 1

and

{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = \frac{\pi }{4}

, then the value of

\left( {a + b} \right) - \left( {\frac{{{a^2} + {b^2}}}{2}} \right) + \left( {\frac{{{a^3} + {b^3}}}{3}} \right) - \left( {\frac{{{a^4} + {b^4}}}{4}} \right) + \,......\,

. is

View answer and explanation

Correct option: A

{\log _e}2

Option A (

{\log _e}2

) is correct because the given condition leads to (1+a)(1+b) = 2, and the series can be expressed as log(1+a) + log(1+b) = log(2).

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)