Question

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

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

Answer 1

$\tan(a+b)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$
$\implies\tan(\sin^{-1}x+\cos^{-1}y)=\dfrac{\tan(\sin^{-1}x)+\tan(\cos^{-1}y)}{1-\tan(\sin^{-1}x)\tan(\cos^{-1}y)}$

If $\theta=\sin^{-1}x\implies\sin\theta=x$, then

$\tan(\sin^{-1}x)=\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

and we know from the Pythagorean identity that $\cos\theta=\sqrt{1-\sin^2\theta}$, so that

$\tan(\sin^{-1}x)=\dfrac x{\sqrt{1-x^2}}$

Similarly, if we let $\varphi=\cos^{-1}y\implies\cos\varphi=y$, we have

$\tan(\cos^{-1}y)=\tan\varphi=\dfrac{\sin\varphi}{\cos\varphi}=\dfrac{\sqrt{1-y^2}}y$

So,

$\tan(\sin^{-1}x+\cos^{-1}y)=\dfrac{\frac x{\sqrt{1-x^2}}+\frac{\sqrt{1-y^2}}y}{1-\frac{x\sqrt{1-y^2}}{y\sqrt{1-x^2}}}$
$\tan(\sin^{-1}x+\cos^{-1}y)=\dfrac{xy+\sqrt{1-x^2}\sqrt{1-y^2}}{y\sqrt{1-x^2}-x\sqrt{1-y^2}}$