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

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 ${\displaystyle \arcsin \left(u\right)}$ 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, ${\displaystyle \cos \left(\arcsin \left(u\right)\right)}$ 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 ${\displaystyle \arctan \left(v\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (1,v)}$. Therefore, ${\displaystyle \cos \left(\arctan \left(v\right)\right)}$ 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.

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

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 ${\displaystyle \arctan \left(v\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (1,v)}$. Therefore, ${\displaystyle \sin \left(\arctan \left(v\right)\right)}$ 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.

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