# If tan alpha=x+1 & tan bita=x-1 Then find what is 2cot(alpha-bita)=?

May 13, 2018

$\rightarrow 2 \cot \left(\alpha - \beta\right) = {x}^{2}$

#### Explanation:

Given that, $\tan \alpha = x + 1 \mathmr{and} \tan \beta = x - 1$.

$\rightarrow 2 \cot \left(\alpha - \beta\right)$

$= \frac{2}{\tan \left(\alpha - \beta\right)} = \frac{2}{\frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \cdot \tan \beta}} = 2 \left[\frac{1 + \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}\right]$

$= 2 \left[\frac{1 + \left(x + 1\right) \cdot \left(x - 1\right)}{\left(x + 1\right) - \left(x - 1\right)}\right]$

$= 2 \left[\frac{\cancel{1} + {x}^{2} \cancel{- 1}}{\cancel{x} + 1 \cancel{- x} + 1}\right] = 2 \left[{x}^{2} / 2\right] = {x}^{2}$

May 13, 2018

$2 \cot \left(\alpha - \beta\right) = {x}^{2}$

#### Explanation:

We have $\tan \alpha = x + 1$ and $\tan \beta = x - 1$

As $\tan \left(\alpha - \beta\right) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

$2 \cot \left(\alpha - \beta\right) = \frac{2}{\tan} \left(\alpha - \beta\right) = 2 \left[\frac{1 + \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}\right]$

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

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

= $\frac{2 {x}^{2}}{2} = {x}^{2}$