JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

x \in \left( {\; - \infty ,\;1} \right)

, then

{\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)

equals

View answer and explanation

Correct option: C

\pi + 2{\tan ^{ - 1}}x

Option C (

\pi + 2{\tan ^{ - 1}}x

) is correct because when

x \in ( - \infty , - 1)

, the expression

{\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)

simplifies to

\pi + 2{\tan ^{ - 1}}x

due to the range of the arctangent function.

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\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)