SATHEE: Inverse Circular Functions 3 Question 6

Inverse Circular Functions 3 Question 6

6. 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) $\frac{3 x + x^{2}}{1 - 3 x^{2}}$

(d) $\frac{3 x + x^{3}}{1 + 3 x^{2}}$

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Answer:

Correct Answer: 6. (a)

Solution:

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

$\begin{aligned} = \tan^{- 1} \frac{x - x^{3} + 2 x}{1 - x^{2} - 2 x^{2}} \\ \Rightarrow \tan^{- 1} y = \tan^{- 1} \frac{3 x - x^{3}}{1 - 3 x^{2}} \\ y = \frac{3 x - x^{3}}{1 - 3 x^{2}} \\ | x | < \frac{1}{\sqrt{3}} \\ \Rightarrow - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \end{aligned}$

$\begin{array}{r} where | x | < \frac{1}{\sqrt{3}} \Rightarrow \tan^{- 1} y = \tan^{- 1} \frac{x + \frac{2 x}{1 - x^{2}}}{1 - x \frac{2 x}{1 - x^{2}}} \\ ∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \frac{x + y}{1 - x y}, \\ where x > 0, y > 0 and x y < 1 \end{array}$

Let $x = \tan θ$

$\begin{array}{lc} \Rightarrow & - \frac{π}{6} < θ < \frac{π}{6} \\ ∴ & \tan^{- 1} y = θ + \tan^{- 1} (\tan 2 θ) = θ + 2 θ = 3 θ \\ \Rightarrow & y = \tan 3 θ \\ \Rightarrow & y = \frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ} \\ \Rightarrow & y = \frac{3 x - x^{3}}{1 - 3 x^{2}} \end{array}$

Inverse Circular Functions 3 Question 6

6. 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

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Inverse Circular Functions 3 Question 6

6. If tan−1⁡y=tan−1⁡x+tan−1⁡2x1−x2, where |x|<13. Then, the value of y is

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6. 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