Inverse Circular Functions Ques 26
- If $\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}$, where $|x|<\frac{1}{\sqrt{3}}$. Then, the value of $y$ is
(a) $\frac{3 x-x^{3}}{1-3 x^{2}}$
(b) $\frac{3 x+x^{2}}{1-3 x^{2}}$
(c) $\frac{3 x-x^{3}}{1+3 x^{2}}$
(d) $\frac{3 x+x^{3}}{1+3 x^{2}}$
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Answer:
Correct Answer: 26.(a)
Solution:
Formula:
Identities of Addition and Substraction:
- Given, $\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} (\frac{2 x}{1-x^{2}})$
$\text { where }|x|<\frac{1}{\sqrt{3}} \Rightarrow \tan ^{-1} y=\tan ^{-1} \{ \frac{x+\frac{2 x}{1-x^{2}}}{1-x \frac{2 x}{1-x^{2}}} \} $
$[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} (\frac{x+y}{1-x y}), $
$\text { where } x>0, y>0 \text { and } x y<1]$
$=\tan ^{-1} (\frac{x-x^{3}+2 x}{1-x^{2}-2 x^{2}}) $
$ \Rightarrow \quad \tan ^{-1} y=\tan ^{-1} (\frac{3 x-x^{3}}{1-3 x^{2}}) $
$ y=\frac{3 x-x^{3}}{1-3 x^{2}} $
$|x|<\frac{1}{\sqrt{3}} $
$ \Rightarrow \quad-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$
Let $x=\tan \theta$
$ \begin{array}{lc} \Rightarrow & -\frac{\pi}{6}<\theta<\frac{\pi}{6} \\ \therefore & \tan ^{-1} y=\theta+\tan ^{-1}(\tan 2 \theta)=\theta+2 \theta=3 \theta \\ \Rightarrow & y=\tan 3 \theta \\ \Rightarrow & y=\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta} \\ \Rightarrow & y=\frac{3 x-x^{3}}{1-3 x^{2}} \end{array} $
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