Let ^-1y=^-1x+^-1(2x1-x^2), where |x|

Question

Let $	an^{-1}y=	an^{-1}x+	an^{-1}\left(\dfrac{2x}{1-x^2}ight)$, where $|x|<\dfrac{1}{\sqrt3}$. Then a value of $y$ is :

Accepted Answer

Given,

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

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

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

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

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

Solutions · 1