JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

{\tan ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4},

then

x

is equal to

View answer and explanation

Correct option: C

\pm \sqrt {\frac{5}{2}}

Option C (

\pm \sqrt {\frac{5}{2}}

) is correct because solving the given inverse tangent equation using the tangent addition formula results in

x = \pm \sqrt {\frac{5}{2}}

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x

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

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