JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

0 < x < 1

, then

\sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)\right.\right.

\left.\left.+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{1 / 2}

is equal to

View answer and explanation

Correct option: C

x \sqrt{1+x^2}

Option C (

x \sqrt{1+x^2}

) is correct because by substituting cot^-1(x) with theta and simplifying the expression using trigonometric identities, the final result is x*sqrt(1+x²).

More Questions from INVERSE TRIGONOMETRY

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The value of

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

is- The value of

\sin \left(2 \tan ^{-1}(1 / 3)\right)+

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

is Find

\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

, then

x

is equal to If

{x^2} + {y^2} + {z^2} = {r^2},

then

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

is equal to

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

is equal to