QJEEM15Ma2506: JEE Main 2015 Mathematics - Class 12 JEEM15Ma2506 Solution: Let where Then a value of is :

Question

Verified Solution • 2025–26 Session moderate

Let

{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right),

where

\left| x \right| &lt; {1 \over {\sqrt 3 }}.

Then a value of

y

is :

Accepted Answer

Given,

{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)

\Rightarrow {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} \right)}}} \right)

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\tan ^{ - 1}}\left( {{{x - {x^3} + 2x} \over {1 - {x^2} - 2{x^2}}}} \right)

\therefore

\,\,\,

{\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{3x - {x^2}} \over {1 - 3{x^2}}}} \right)

\Rightarrow y = {{3x - {x^3}} \over {1 - 3{x^2}}}

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