Answer:
[xy + √(1−x²) √(1−y²)] / [y √(1−x²) − x √(1−y²)]
Explanation:
tan(sin⁻¹x + cos⁻¹y)
Use angle sum formula:
[tan(sin⁻¹x) + tan(cos⁻¹y)] / [1 − tan(sin⁻¹x) tan(cos⁻¹y)]
To evaluate these expressions, I suggest drawing right triangles.
For example, let's draw a triangle where x is the side opposite of angle θ, and the hypotenuse is 1. Therefore:
sin θ = x/1
θ = sin⁻¹x
Using Pythagorean theorem, the adjacent side is √(1−x²). Therefore:
tan θ = x / √(1−x²)
tan(sin⁻¹x) = x / √(1−x²)
Draw a new triangle. This time we'll make y the adjacent side to angle θ.
cos θ = y/1
θ = cos⁻¹y
Using Pythagorean theorem, the opposite side is √(1−y²). Therefore:
tan θ = √(1−y²) / y
tan(cos⁻¹y) = √(1−y²) / y
Substituting:
[x / √(1−x²) + √(1−y²) / y] / [1 − x / √(1−x²) × √(1−y²) / y]
Multiply top and bottom by √(1−x²).
[x + √(1−x²) √(1−y²) / y] / [√(1−x²) − x × √(1−y²) / y]
Multiply top and bottom by y.
[xy + √(1−x²) √(1−y²)] / [y √(1−x²) − x √(1−y²)]