(a) Solve the following equation for x.tan−1 1−x1+x=12tan−1 x;x>0(b) Find the value of tan tan-1 xy−tan−1 x−yx+y

Given, tan−1 1−x1+x=12tan−1 x;x>0⇒ 2tan−1 1−x1+x=tan−1 x[multiplying both sides by 2]⇒ tan-121−x1+x1−1−x1+x2=tan−1 x⇒∵2tan−1 x=tan−1 2x1−x2;−1 0given, so we do not take x=−13∴x=13 is the only solution of the given equation.b) We know that tan−1 x−tan−1 y=tan−1 x−y1+xyReplacing x by xy and y by x−yx+ytan−1 xy−tan−1 x−yx+y=tan−1 xy−x−yx+y1+xyx−yx+y=tan−1 x(x+y)−y(x−y)y(x+y)y(x+y)+x(x−y)y(x+y)=tan−1 x(x+y)−y(x−y)y(x+y)×y(x+y)y(x+y)+x(x−y)=tan−1 x(x+y)−y(x−y)y(x+y)+x(x+y)×y(x+y)y(x+y)=tan−1 x(x+y)−y(x−y)y(x+y)+x(x+y)=tan−1 x2+xy−yx+y2yx+y2+x2−xy=tan−1 x2+y2+yx−yxy2+x2+yx−xy=tan−1 x2+y2y2+x2=tan−1 (1)=tan−1 tan π4 As, tan Π4=1=π4

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(a) Solve the following equation for x.
$\tan^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} \tan^{- 1} x; x > 0$
(b) Find the value of tan $\tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{x - y}{x + y})$

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Detailed Solution

$Given, \tan^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} \tan^{- 1} x; x > 0$
$\Rightarrow 2 \tan^{- 1} (\frac{1 - x}{1 + x}) = \tan^{- 1} x$
[multiplying both sides by 2]
$\begin{array}{l} \Rightarrow \tan^{- 1} [\frac{2 (\frac{1 - x}{1 + x})}{1 - {(\frac{1 - x}{1 + x})}^{2}}] = \tan^{- 1} x \\ \Rightarrow [∵ 2 \tan^{- 1} x = \tan^{- 1} (\frac{2 x}{1 - x^{2}}); - 1 < x < 1] \\ \Rightarrow \tan^{- 1} [\frac{2 (1 - x) (1 + x)}{(1 + x)^{2} - (1 - x)^{2}}] = \tan^{- 1} x \\ \Rightarrow \tan^{- 1} (\frac{2 (1 - x^{2})}{4 x}) = \tan^{- 1} x \\ [∵ (a - b) (a + b) = a^{2} - b^{2} \\ and (a + b)^{2} - (a - b)^{2} = 4 ab] \end{array}$
$\begin{array}{l} \Rightarrow \frac{2 (1 - x^{2})}{4 x} = x \\ \Rightarrow \frac{1 - x^{2}}{2 x} = x \Rightarrow 1 - x^{2} = 2 x^{2} \\ \Rightarrow 3 x^{2} = 1 \Rightarrow x^{2} = \frac{1}{3} \\ \Rightarrow x = \pm \frac{1}{\sqrt{3}} \end{array}$
$[[∵ x > 0 given, so we do not take x = - \frac{1}{\sqrt{3}}]$
$[∴ x = \frac{1}{\sqrt{3}} is the only solution of the given equation.]$

b) We know that $\tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + xy})$
Replacing x by $\frac{x}{y} and y by \frac{x - y}{x + y}$
$\tan^{- 1} \frac{x}{y} - \tan^{- 1} (\frac{x - y}{x + y}) = \tan^{- 1} [\frac{\frac{x}{y} - \frac{x - y}{x + y}}{1 + (\frac{x}{y}) (\frac{x - y}{x + y})}]$
$\begin{array}{l} = \tan^{- 1} (\frac{\frac{x (x + y) - y (x - y)}{y (x + y)}}{\frac{y (x + y) + x (x - y)}{y (x + y)}}) \\ = \tan^{- 1} (\frac{x (x + y) - y (x - y)}{y (x + y)} \times \frac{y (x + y)}{y (x + y) + x (x - y)}) \\ = \tan^{- 1} (\frac{x (x + y) - y (x - y)}{y (x + y) + x (x + y)} \times \frac{y (x + y)}{y (x + y)}) \\ = \tan^{- 1} (\frac{x (x + y) - y (x - y)}{y (x + y) + x (x + y)}) \\ = \tan^{- 1} (\frac{x^{2} + xy - yx + y^{2}}{yx + y^{2} + x^{2} - xy)}) \\ = \tan^{- 1} (\frac{x^{2} + y^{2} + yx - yx}{y^{2} + x^{2} + yx - xy)}) \\ = \tan^{- 1} (\frac{x^{2} + y^{2}}{y^{2} + x^{2}}) \\ = \tan^{- 1} (1) \end{array}$
$\begin{array}{l} = \tan^{- 1} (\tan \frac{π}{4}) As, \tan \frac{Π}{4} = 1 \\ = \frac{π}{4} \end{array}$