JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

\sin ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4} \ldots \ldots \infty\right)+\cos ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4} \ldots . \infty\right)=\frac{\pi}{2}

for

0 < |x| < \sqrt{2}

, then

\cot ^{-1} x

equals to

View answer and explanation

Correct option: D

\frac{\pi}{4}

Option D (

\frac{\pi}{4}

) is correct because solving the given equation after simplifying the infinite geometric series leads to

x=1

, and thus

\cot^{-1}(1) = \frac{\pi}{4}

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)