# cot^-1x + cot^-1(1+x)=pi/4?

May 10, 2018

$x = - 1 , 2$

#### Explanation:

${\cot}^{- 1} \left(x\right) + {\cot}^{- 1} \left(1 + x\right) = \frac{\pi}{4}$

Take $\tan$ of both sides:

$\tan \left[{\cot}^{- 1} \left(x\right) + {\cot}^{- 1} \left(1 + x\right)\right] = \tan \left(\frac{\pi}{4}\right)$

Expand the left side using the sum formula:

$\tan \left(x + y\right) = \frac{\tan \left(x\right) + \tan \left(y\right)}{1 - \tan \left(x\right) \tan \left(y\right)}$

So in this case:

$\frac{\tan \left({\cot}^{- 1} \left(x\right)\right) + \tan \left({\cot}^{- 1} \left(1 + x\right)\right)}{1 - \tan \left({\cot}^{- 1} \left(x\right)\right) \cdot \tan \left({\cot}^{- 1} \left(1 + x\right)\right)} = 1$

Simplify:

$\frac{\frac{1}{x} + \frac{1}{x + 1}}{1 - \frac{1}{x} \cdot \frac{1}{x + 1}} = 1$

$\frac{x + 1 + x}{x \left(x + 1\right)} = 1 - \frac{1}{{x}^{2} + x}$

$\frac{2 x + 1}{{x}^{2} + x} = \frac{{x}^{2} + x - 1}{{x}^{2} + x}$

${x}^{2} - x - 2 = 0$

$\left(x + 1\right) \left(x - 2\right) = 0$

$x = - 1 , 2$