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

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

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

Combine and simplify the denominator.

The functions cosine and arccosine are inverses.

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

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

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

Combine and simplify the denominator.

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

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 ${\displaystyle b=v}$.

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