# Question #e9266

Aug 29, 2017

After using u= ${\cot}^{- 1} \left(\frac{\sqrt{1 + x}}{\sqrt{1 - x}}\right)$ transform,

$\cot u = \frac{\sqrt{1 + x}}{\sqrt{1 - x}}$

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

$\left(1 - x\right) \cdot {\left(\cot u\right)}^{2} = 1 + x$

${\left(\cot u\right)}^{2} - x \cdot {\left(\cot u\right)}^{2} = 1 + x$

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

$x \cdot {\left(\csc u\right)}^{2} = {\left(\csc u\right)}^{2} \cdot \left[{\left(\cos u\right)}^{2} - {\left(\sin u\right)}^{2}\right]$

$x = {\left(\cos u\right)}^{2} - {\left(\sin u\right)}^{2}$

$x = \cos 2 u$

Hence,

$y = {\left(\sin \left[{\cot}^{- 1} \left(\frac{\sqrt{1 + x}}{\sqrt{1 - x}}\right)\right]\right)}^{2}$

=${\left(\sin u\right)}^{2}$

=$\frac{1}{2} \cdot 2 {\left(\sin u\right)}^{2}$

=$\frac{1}{2} \cdot \left(1 - \cos 2 u\right)$

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

#### Explanation:

1) I used $u = {\cot}^{- 1} \left(\frac{\sqrt{1 + x}}{\sqrt{1 - x}}\right)$ transform for finding x in terms of u. Finally I found $x = \cos 2 u$.

2) I rewrote ${\left(\sin u\right)}^{2}$ in terms of $\cos 2 u$. Finally I found ${\left(\sin \left[{\cot}^{- 1} \left(\frac{\sqrt{1 + x}}{\sqrt{1 - x}}\right)\right]\right)}^{2}$ as $\frac{1 - x}{2}$.