Expand the Trigonometric Expression sin(arcsin(u)+arctan(v))

The functions sine and arcsine are inverses.

Draw a triangle in the plane with vertices ${\displaystyle (1,v)}$, ${\displaystyle (1,0)}$, and the origin. Then ${\displaystyle {\tan }^{-1}\left(v\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (1,v)}$. Therefore, ${\displaystyle \cos \left({\tan }^{-1}\left(v\right)\right)}$ is ${\displaystyle \frac{1}{\sqrt{1+v^2}}}$.

Multiply ${\displaystyle \frac{1}{\sqrt{1+v^2}}}$ by ${\displaystyle \frac{\sqrt{1+v^2}}{\sqrt{1+v^2}}}$.

Combine and simplify the denominator.

Combine ${\displaystyle u}$ and ${\displaystyle \frac{\sqrt{1+v^2}}{1+v^2}}$.

Draw a triangle in the plane with vertices ${\displaystyle (\sqrt{1^2-u^2},u)}$, ${\displaystyle (\sqrt{1^2-u^2},0)}$, and the origin. Then ${\displaystyle {\sin }^{-1}\left(u\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (\sqrt{1^2-u^2},u)}$. Therefore, ${\displaystyle \cos \left({\sin }^{-1}\left(u\right)\right)}$ is ${\displaystyle \sqrt{1-u^2}}$.

Rewrite ${\displaystyle 1}$ as ${\displaystyle 1^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=1}$ and ${\displaystyle b=u}$.

Draw a triangle in the plane with vertices ${\displaystyle (1,v)}$, ${\displaystyle (1,0)}$, and the origin. Then ${\displaystyle {\tan }^{-1}\left(v\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (1,v)}$. Therefore, ${\displaystyle \sin \left({\tan }^{-1}\left(v\right)\right)}$ is ${\displaystyle \frac{v}{\sqrt{1+v^2}}}$.

Multiply ${\displaystyle \frac{v}{\sqrt{1+v^2}}}$ by ${\displaystyle \frac{\sqrt{1+v^2}}{\sqrt{1+v^2}}}$.

Combine and simplify the denominator.

Multiply ${\displaystyle \sqrt{(1+u)(1-u)}\frac{v\sqrt{1+v^2}}{1+v^2}}$.