Question

Write the given expression in terms of x and y only.

tan(sin^-1(x)+cos^-1(y))

Answer 1

`\tan(a+b)=(\tan a+\tan b)/(1-\tan a\tan b)`

`\implies\tan(\sin^(-1)x+\cos^(-1)y)=(\tan(\sin^(-1)x)+\tan(\cos^(-1)y))/(1-\tan(\sin^(-1)x)\tan(\cos^(-1)y))`

If
`\theta=\sin^(-1)x\implies\sin\theta=x`, then


`\tan(\sin^(-1)x)=\tan\theta=(\sin\theta)/(\cos\theta)`

and we know from the Pythagorean identity that
`\cos\theta=√(1-\sin^2\theta)`, so that


`\tan(\sin^(-1)x)=\frac x{√(1-x^2)}`

Similarly, if we let
`\varphi=\cos^(-1)y\implies\cos\varphi=y`, we have


`\tan(\cos^(-1)y)=\tan\varphi=(\sin\varphi)/(\cos\varphi)=\frac{√(1-y^2)}y`

So,


`\tan(\sin^(-1)x+\cos^(-1)y)=\frac{\frac x{√(1-x^2)}+\frac{√(1-y^2)}y}{1-(x√(1-y^2))/(y√(1-x^2))}`

`\tan(\sin^(-1)x+\cos^(-1)y)=(xy+√(1-x^2)√(1-y^2))/(y√(1-x^2)-x√(1-y^2))`