Question

How to solve tan(arcsin(x) + arccos(y)) in terms of x and y only?

I do not understand how to plug in the values I get of: `sqrt(1-x^2)` and `sqrt(1-y^2)` into tan(x+y)

Answer 1

`[(xy+sqrt(1-x^2-y^2+x^2y^2))/(ysqrt(1-x^2)-xsqrt(1-y^2))]`

Explanation:

To use the formula for inverse function, we assume that, `0< x,y <1and(x/sqrt(1-x^2))(sqrt(1-y^2)/y)<1`
We know that,
`sin^-1x+cos^-1y=tan^-1(x/sqrt(1-x^2))+tan^-1(sqrt(1-y^2)/y)=tan^-1[((x/sqrt(1-x^2))+(sqrt(1-y^2)/y))/(1-(x/sqrt(1-x^2))(sqrt(1-y^2)/y))]`
`=tan^-1[(xy+sqrt(1-x^2)sqrt(1-y^2))/(ysqrt(1-x^2)-xsqrt(1-y^2))]`
`=tan^-1[(xy+sqrt(1-x^2-y^2+x^2y^2))/(ysqrt(1-x^2)-xsqrt(1-y^2))]`
Hence,
`tan(sin^-1x+cos^-1y)=tan(tan^-1[(xy+sqrt(1-x^2-y^2+x^2y^2))/(ysqrt(1-x^2)-xsqrt(1-y^2))])`
`=[(xy+sqrt(1-x^2-y^2+x^2y^2))/(ysqrt(1-x^2)-xsqrt(1-y^2))]`