<s>[INST] Simplify $\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})}.$ First I denote $x=\arctan{2}$ and $y=\arcsin{\frac{1}{\sqrt{10}}}$ and then use the addition formula for sine:
$$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}=\sin{x}\cos{y}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Now I use the fact that $\cos{y}=1-\sin^2{y}$ which gives
$$\cos{y}=1-\sin^2{\left(\arcsin{\frac{1}{\sqrt{10}}}\right)}=1-\frac{1}{10}=\frac{9}{10}$$
So I have reduced the problem to the following
$$\sin{x}\cdot \frac{9}{10}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Here I'm stuck. I cant just divide this expression by $(\sin{x}\cos{x})$ because I'd still be left with sine and cos terms and I'd also change the value of the expression. Everything now boils down to compute $\sin(\arctan{2})$ and $\cos{\arctan2}.$ I also tried rewriting $\cos(\arctan{2})$ as $1-\sin^2(\arctan{2}),$ but to no avail.
Any suggestions on 


*

*how to proceed from where I left;

*how tocompute this by means more effective;

*both of the above.

 [/INST] Hint:
$$\sin x =\sqrt{1-\cos^2 x} =\sqrt{1-\frac{1}{1+\tan^2 x}} $$
$$\cos x =\sqrt{\frac{1}{1+\tan^2 x}} $$
Since 
$$ 1+\tan^2 x =1+\frac{\sin x^2}{\cos^2 x} = \frac{\cos^2x+\sin x^2}{\cos^2 x} =\frac{1}{\cos^2 x}$$
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