Let  \tan ^{-1}y=\tan ^{-1}x\, +\, \tan ^{-1}\left ( \frac{2x}{1-x^{2}} \right ),    where \left | x \right |< \frac{1}{\sqrt{3}}.  

Then a value of y  is :

  • Option 1)

    \frac{3x-x^{3}}{1-3x^{2}}

  • Option 2)

    \frac{3x+x^{3}}{1-3x^{2}}

  • Option 3)

    \frac{3x-x^{3}}{1+3x^{2}}

  • Option 4)

    \frac{3x+x^{3}}{1+3x^{2}}

 

Answers (1)

As we learned in 

Formulae of Inverse Trigonometric Functions -

\cos ^{-1}x + \cos ^{-1}y = \cos ^{-1}\left \left ( xy- \sqrt{1-x^{2}}\sqrt{1-y^{2}} \right )

- wherein

x\geqslant 0, y\geqslant 0

 

\left | x \right |< \frac{1}{\sqrt{3}}\Rightarrow \tan ^{-1}x< 30^{\circ} \: , \tan^{-1}\frac{2x}{1-x^{2}}< 60^{\circ}

\Rightarrow \tan ^{-1}y< 90^{\circ}

\tan ^{-1}y= \tan ^{-1}x+\tan ^{-1}\frac{2x}{1-x^{2}}

              =\tan ^{-1}x+2\tan ^{-1}x

              =3\tan ^{-1}x

y=tan(3tan^{-1}x)

    =\frac{3tan(tan^{-1}x)-\left [ tan(tan^{-3}x) \right ]}{1-3(tan \: tan^{-1}x)^{2}}

    =\frac{3x-x^{3}}{1-3x^{2}}                                                            \left [ \because tan \: 3x=\frac{3tanx-tan^{3}x}{1-3tan^{2}x} \right ]

 


Option 1)

\frac{3x-x^{3}}{1-3x^{2}}

This option is correct

Option 2)

\frac{3x+x^{3}}{1-3x^{2}}

This option is incorrect

Option 3)

\frac{3x-x^{3}}{1+3x^{2}}

This option is incorrect

Option 4)

\frac{3x+x^{3}}{1+3x^{2}}

This option is incorrect

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