AQA MPC3 June 2015 Q8 please help

Hi all I have worked on this question and cannot see how to achieve the final step could someone please help me from where I have got up to?

I started off changing ID to cot^2x= K-4/3

I then inverted to get tan^2x = 3/K-3 to get the K-4 as denominator but how do you get the K-1 part? I saw I can change the tan^2x to sec^2x +1 but I am a K missing!

$\displaystyle \frac{3}{k-4}+1 = \frac{k-1}{k-4}$.

Hi thankyou for the reply yea its this bit I just cant see how does K-1 form please forgive me I cant quite get it from this.

Hi thankyou for the reply yea its this bit I just cant see how does K-1 form please forgive me I cant quite get it from this.

3/(k-4) + 1 = 3/(k-4) + (k-4)/(k-4), since anything divided by itself is 1.
Now since the fractions have the same denominators, we can just add the numerators to get (3+k-4)/(k-4) = (k-1)/(k-4),
as required.