Attempt: I've used that $\sin(x) = (2t)/(1+t^2)$ and $\cos(x) = (1-t^2)/(1+t^2)$. However I don't seem to get a perfect square, instead I get $$(2/((1+t^2)^2))(14t^2 +20t + 38)$$.
I'm not sure if there error is with my method or my workings.
Any help would be greatly appreciated.
$$\begin{align} 2(1+\cos(x))(5\sin(x)+12\cos(x)+13)&=2\left(1+\frac{1-t^2}{1+t^2}\right)\left(\frac{10t}{1+t^2}+\frac{12-12t^2}{1+t^2}+13\right)\\ &=\frac{4}{(1+t^2)^2}(t+5)^2 \end{align}$$