If $x > 0, y > 0$ and $x>y$, then $\tan ^{-1}(x / y)+\tan ^{-1}[(x+y) /(x-y)]$ is equal to

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Medium

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Application +4 | -1

x > 0, y > 0

and

x>y

, then

\tan ^{-1}(x / y)+\tan ^{-1}[(x+y) /(x-y)]

is equal to

View answer and explanation

Correct option: C

3 \pi / 4

Option C (

3 \pi / 4

) is correct because the expression simplifies to

\pi + \tan^{-1}(-1)

, which equals

3\pi/4

given the condition x > y > 0.

Practice All Chapter MCQs →

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\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)