Rewrite $\sin(\cos^{-1}(x)-\tan^{-1}(y))$ as an algebraic function of $x$ and $y$. Rewrite the expression as an algebraic function of $x$ and $y$:
$$\sin(\cos^{-1}(x)-\tan^{-1}(y)).$$
I am unsure of how to change this into an algebraic function, yet I am able to simplify inso sin and cos.
thank you
 A: Let $\theta = \arccos x$.  Then $\cos\theta = x$, so we can draw a right triangle with adjacent side of length $|x|$ and hypotenuse of length $1$.  By the Pythagorean Theorem, the opposite side has length $\sqrt{1 - x^2}$.   
Let $\varphi = \arctan y$.  Then $\tan\varphi = y$, so we can draw a right triangle with opposite side of length $|y|$ and adjacent side of length $1$.  By the Pythagorean Theorem, the hypotenuse has length $\sqrt{1 + y^2}$.  
Thus,
\begin{align*}
\sin(\arccos x + \arctan y) & = \sin(\theta + \varphi)\\
                            & = \sin\theta\cos\varphi + \cos\theta\sin\varphi
\end{align*}
From the diagrams, we see that 
\begin{align*}
\cos\theta & = x\\
\sin\theta & = \sqrt{1 - x^2}\\
\cos\varphi & = \frac{1}{\sqrt{1 + y^2}}\\
\sin\varphi & = \frac{y}{\sqrt{1 + y^2}}
\end{align*}
Make the appropriate substitutions, then simplify.
A: 


Now we need to apply the double angle formula to get the expression in the correct form: 
$$\sin(\cos^{-1}(x)-\tan^{-1}(y))$$ = $$\sin(\cos^{-1}(x)\cos(\tan^{-1}(y)-\cos(\cos^{-1}(x)\sin(\tan^{-1}(y)))$$=$$\sin(\cos^{-1}(x)\cos(\tan^{-1}(y)-x\sin(\tan^{-1}(y))).$$  
Now it's a matter of substitution and rearrangement. I'll leave you to the tedious details. 
A: Let $\cos^{-1}x=A\implies\cos A=x,\tan^{-1}y=B\implies \tan B=y$
Using the definition of Principal values of Inverse trigonometric functions,
$0\le A\le\pi$ and $-\dfrac\pi2\le B\le \dfrac\pi2$
$\implies\sin A\ge0\implies\sin A=+\sqrt{1-\cos^2A}=+\sqrt{1-x^2}$
and $\cos B\ge0\implies\cos B=+\dfrac1{\sqrt{1+\tan^2B}}=\dfrac1{\sqrt{1+y^2}}$
and consequently, $\sin B=\tan B\cdot\cos B=\cdots$
Finally apply $\sin(A-B)$ formula.
A: $$\sin(\cos^{-1}(x)-\tan^{-1}(y))=\sin(\cos^{-1}(x))\cos(\tan^{-1}(y))-\cos(\cos^{-1}(x))\sin(\tan^{-1}(y))=\frac{\sqrt{1-x^2}-xy}{\sqrt{1+y^2}}.$$
Indeed, 
$$\sin(\cos^{-1}(x))=\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-x^2},$$
and
$$\cos(\tan^{-1}(y))=\cos\phi=\frac1{\sqrt{1+\tan^2\phi}}=\frac1{\sqrt{1+y^2}},\\
\sin(\tan^{-1}(y))=\sin\phi=\frac{\tan\phi}{\sqrt{1+\tan^2\phi}}=\frac y{\sqrt{1+y^2}}.$$
