If $y=\tan ^{-1} \frac{1-x}{1+x}, 0 \leq x \leq 1$, then

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

y=\tan ^{-1} \frac{1-x}{1+x}, 0 \leq x \leq 1

, then

View answer and explanation

Correct option: B

0 \leq y \leq \pi / 4

The correct answer is B (

0 \leq y \leq \pi / 4

) because substituting

x = \tan \theta

and simplifying the expression leads to this range for y.

Practice All Chapter MCQs →

\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)