+118608 I have not got the desired answer, perhaps another mathematician can find my error ?? Thanks :) 
Also Kath, this was a super hard one, make sure you can do the easy ones first.(I know that you are struggling with these) 
Reduce expression to single function
You cannot express as a single function because there is an equal sign.
I assume you want me to prove that the LHS=RHS.
\(\frac{sin(3\theta)}{sin(\theta)}+\frac{cos(3\theta)}{cos(\theta)}=2cot(2\theta)\)
You need to memorise that:
sin(A+B)=sinAcosB+cosAsinB
and
cos(A+B)=cosAcosB - sinAsinB
so
\(sin(3\theta)\\ =sin(2\theta+\theta)\\ =sin(2\theta)cos(\theta)+cos(2sin\theta)sin(\theta)\\ and\\ cos(3\theta)\\ =cos(2\theta+\theta)\\ =cos(2\theta)cos(\theta) - sin(2\theta)sin(\theta)\\\)
\(\frac{sin(3\theta)}{sin(\theta)}+\frac{cos(3\theta)}{cos(\theta)}=2cot(2\theta)\)
LHS
\(=\frac{sin(3\theta)}{sin(\theta)}+\frac{cos(3\theta)}{cos(\theta)}\\ =\frac{sin(2\theta)cos(\theta)+cos(2\theta)sin(\theta)}{sin(\theta)}+\frac{cos(2\theta)cos(\theta) - sin(2\theta)sin(\theta)}{cos(\theta)}\\ =\frac{sin(2\theta)cos(\theta)}{sin(\theta)}+\frac{cos(2\theta)sin(\theta)}{sin(\theta)} +\frac{cos(2\theta)cos(\theta)}{cos(\theta)} - \frac{sin(2\theta)sin(\theta)}{cos(\theta)}\\ =\frac{sin(2\theta)cos(\theta)}{sin(\theta)}+\frac{cos(2\theta)}{1} +\frac{cos(2\theta)}{1} - \frac{sin(2\theta)sin(\theta)}{cos(\theta)}\\ =\frac{sin(2\theta)cos(\theta)}{sin(\theta)} - \frac{sin(2\theta)sin(\theta)}{cos(\theta)}+\frac{2cos(2\theta)}{1}\\ \)
\(=\frac{sin(2\theta)}{1}\left[\frac{cos(\theta)}{sin(\theta)} - \frac{sin(\theta)}{cos(\theta)}\right]+\frac{2cos(2\theta)}{1}\\ =\frac{sin(2\theta)}{1}\left[\frac{cos^2(\theta)}{sin(\theta)cos(\theta)} - \frac{sin^2(\theta)}{sin(\theta)cos(\theta)}\right] +\frac{2cos(2\theta)}{1}\\ =\frac{sin(2\theta)}{1}\left[\frac{cos^2(\theta)-sin^2(\theta)}{sin(\theta)cos(\theta)} \right] +\frac{2cos(2\theta)}{1}\\ =\frac{sin(2\theta)}{1}\left[\frac{cos(2\theta)}{sin(\theta)cos(\theta)} \right] +\frac{2cos(2\theta)}{1}\\ =\frac{sin(2\theta)}{1}\times2\left[\frac{cos(2\theta)}{2sin(\theta)cos(\theta)} \right] +\frac{2cos(2\theta)}{1}\\ =\frac{sin(2\theta)}{1}\times2\left[\frac{cos(2\theta)}{sin(2\theta)} \right] +\frac{2cos(2\theta)}{1}\\ =\frac{1}{1}\times2\left[\frac{cos(2\theta)}{1} \right]+\frac{2cos(2\theta)}{1}\\ =4cos(2\theta) \)
\(\ne RHS\)
It did not get this as an equality. It is quite likely that i made a mistake. 
+33616 The two sides are not generally equal!
Try a few example values (say 20 and 25 degrees) to check.
(For A read theta).
On the lhs, use the standard trig identities, sin(3A) = 3sin(A) - 4sin(A)^3
and cos(3A) = 4cos(A)^3 - 3cos(A).
On the rhs, cot(2A) = 1/tan(2A), and tan(2A) = 2tan(A)/(1-tan(A)^2).
Also, notice that cos(A)^2 - sin(A)^2 = cos(A)^2(1 - tan(A)^2).
Take care with the first line though, when the sine and cosine are cancelled, you are, in effect, excluding sin(A) = 0 and cos(A) = 0. In fact, sin(A) = 0 and cos(A) = 0 are not allowed.
