JEE MAIN - Mathematics (2015 (Offline) - No. 13)
Let $${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right),$$
where $$\left| x \right| < {1 \over {\sqrt 3 }}.$$ Then a value of $$y$$ is :
where $$\left| x \right| < {1 \over {\sqrt 3 }}.$$ Then a value of $$y$$ is :
$${{3x - {x^3}} \over {1 + 3{x^2}}}$$
$${{3x + {x^3}} \over {1 + 3{x^2}}}$$
$${{3x - {x^3}} \over {1 - 3{x^2}}}$$
$${{3x + {x^3}} \over {1 - 3{x^2}}}$$
คำอธิบาย
Given,
$${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)$$
$$ \Rightarrow {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} \right)}}} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\tan ^{ - 1}}\left( {{{x - {x^3} + 2x} \over {1 - {x^2} - 2{x^2}}}} \right)$$
$$\therefore$$ $$\,\,\,$$ $${\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{3x - {x^2}} \over {1 - 3{x^2}}}} \right)$$
$$ \Rightarrow y = {{3x - {x^3}} \over {1 - 3{x^2}}}$$
$${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)$$
$$ \Rightarrow {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} \right)}}} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\tan ^{ - 1}}\left( {{{x - {x^3} + 2x} \over {1 - {x^2} - 2{x^2}}}} \right)$$
$$\therefore$$ $$\,\,\,$$ $${\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{3x - {x^2}} \over {1 - 3{x^2}}}} \right)$$
$$ \Rightarrow y = {{3x - {x^3}} \over {1 - 3{x^2}}}$$