Interesting problem

(b)

We start with  sqrt(x)  on the display.

Then take the  arctan  of what is on the display.

Then take the  cos  of what is on the display.  This gives us  \(\frac{1}{\sqrt{x+1}}\)

Then...just like from part (a).....we can take the  arctan  of what is on the display.

Then take the  cot  of what is on the display.

Altogether:

\(\begin{array}{ccc} &\cot( \arctan( &\cos( \arctan(\sqrt{x}\,)\,) & ) \quad ) \\~\\ =&\cot( \arctan( &\frac{1}{\sqrt{x+1}}& ) \quad ) \\~\\ =&\sqrt{x+1} \end{array}\)

Check: https://www.wolframalpha.com/input/?i=cot(arctan(cos(arctan(sqrt(x)))))

(c)

Let's define functions  a,  b,  and  c  like this:

a(x)  =  cot( arctan(x) )   =   \(\frac1x\)       where  x  is any real number except  0

b(x)  =  cos( arctan(x) )  =  \(\sqrt{\frac{1}{x^2+1}}\)       where  x  is any real number

c(x)  =  cos( arcsin(x) )   =   \(\sqrt{1-x^2}\)       where  x  is a real number the interval  [-1, 1]

We start with  0  on the display.

Then we can take the  cos  to get  1

Then we can do  b  to get  \(\sqrt{\frac12}\)

Then we can do  b  to get  \(\sqrt{\frac23}\)

Then we can do  c  to get  \(\sqrt{\frac13}\)

Then we can do  b  to get  \(\sqrt{\frac34}\)

Then we can do  c  to get  \(\sqrt{\frac14}\)

Then we can do  b  to get  \(\sqrt{\frac45}\)

Then we can do  b  to get  \(\sqrt{\frac59}\)

Then we can do  a  to get  \(\frac{3}{\sqrt5}\)

Check: https://www.wolframalpha.com/input/?i=cot(arctan(cos(arctan(cos(arctan(. . .

Thanks!

Just wondering, did you use a computer to find this?