Let { an ^{ - 1}}y = { an ^{ - 1}}x + { an ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} ight), where \left| x ight|

Question

Let {	an ^{ - 1}}y = {	an ^{ - 1}}x + {	an ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} ight), where \left| x ight| < {1 \over {\sqrt 3 }}. Then a value of y is :

Accepted Answer

Answer: C. {{3x - {x^3}} \over {1 - 3{x^2}}} Given, {	an ^{ - 1}}y = {	an ^{ - 1}}x + {	an ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} ight) \Rightarrow {	an ^{ - 1}}y = {	an ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} ight)}}} ight) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {	an ^{ - 1}}\left( {{{x - {x^3} + 2x} \over {1 - {x^2} - 2{x^2}}}} ight) 	herefore \,\,\, {	an ^{ - 1}}y = {	an ^{ - 1}}\left( {{{3x - {x^2}} \over {1 - 3{x^2}}}} ight) \Rightarrow y = {{3x - {x^3}} \over {1 - 3{x^2}}}

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 :

Answer

Explanation

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