JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

\tan ^{-1}(x+h)=\tan ^{-1}(x)+(h \sin y)(\sin y)-(h \sin y)^2 \cdot \frac{\sin 2 y}{2}+(h \sin y)^3 \cdot \frac{\sin 3 y}{3}+\ldots \ldots \ldots

Where

x \in(0,1), y \in(\pi / 4, \pi / 2)

, then

View answer and explanation

Correct option: C

y=\cot ^{-1} x

Option C (

y=\cot^{-1} x

) is correct because by comparing the given series with the Taylor series expansion of

\tan^{-1}(x+h)

, we can deduce that

y = \cot^{-1}x

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)