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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)

best_answer

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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