Calculating certain functions if only certain buttons on a calculator are permitted 
A calculator is broken. The only keys that work are  $\sin, \cos, \tan, \cot, \arcsin, \arccos$, and $\arctan$ buttons.
The original display is $0$.
In this problem, we will prove that given any positive rational number $q$, show that pressing some finite sequence of buttons will yield $q$.
Functions are always in radian form.
(a) Find and prove that there exists a  sequence of buttons that will turn  $\sqrt x$ into $\sqrt{x+1}$.

For this, I got lucky and tried out $\sec (\arctan(x))$ on my calculator and got $\sqrt{x^2+1}$, so I just used the reciprocal and got $\cos (\arctan (x)) = \frac{1}{\sqrt{x^2+1}}$. However, how I can actually prove this?

(b) Prove that there exists a sequence of buttons that will yield $\frac{3}{\sqrt{5}}$.

I know that to go from $x$ to $\frac{1}{x}$ it is $\cot (\arctan (x))$, and I will have to use part a) to get to part b).
How do I utilize this? I also know that $\sqrt{\frac{9}{5}}$ = $\frac{3}{\sqrt{5}}$, but after that, I'm stuck. Any hints?
 A: (a) Since $\tan(\arctan(x))= x$, we can invert the equation to get $\frac{1}{\tan(\arctan(x))}=\boxed{\cot(\arctan(x))}$
(b) If you draw a triangle with legs 1 and $\sqrt{x}$, the hypotenuse would be $\sqrt{x+1}$. To have $\frac{1}{\sqrt{x+1}}$, you would do $\cos(\arctan(\sqrt{x}))$. To get $\sqrt{x+1}$ from $\sqrt{x}$, you would need to do $\frac{1}{\cos(\arctan(\sqrt{x}))}$. We know we can get the reciprocal from part (a). So, getting the reciprocal of $\frac{1}{\cos(\arctan(\sqrt{x}))}$, we get $\boxed{\cot(\arctan(\cos(\arctan(\sqrt{x})))}$
(c) Let's call the function we got a part (a) as $A$, and the function we got a part (b) as $B$. Then, starting from 0, we get: $$0 \overset{B}{\to} \sqrt 1 \overset{B}{\to} \sqrt 2 \overset{B}{\to} \sqrt 3 \overset{B}{\to} \sqrt 4 \overset{A}{\to} \sqrt \frac{1}{4} \overset{B}{\to} \sqrt \frac{5}{4} \overset{A}{\to} \sqrt \frac{4}{5} \overset{B}{\to} \sqrt \frac{9}{5} = \frac{3}{\sqrt 5}$$From, this we get the sequence of $\boxed{BBBBABAB}$, which is equal to... Something very long, which I do not need to type!
Thus, I have proved that I can create the number $\frac{3}{\sqrt 5}$ from $0$!
A: For question (b):
Just as MvG said above, we can get $\frac{1}{\sqrt{5}}$. Note that $\sec(\arcsin x)=\frac{1}{\sqrt{1-x^2}}$, so $\sqrt{\frac{5}{4}}$ may be obtained, which follows $\sqrt{\frac{4}{5}}$. Using $\sec(\arctan x)=\sqrt{x^2+1}$ can give the answer. 
