1. ## trig identity

I have problems with solving this identity, this is how I started

$\frac{cos4x+1}{ctgx+tgx} = \frac{1}{2}\cdot\sin4x (1)$

$\frac{cosx(2x+2x)+1}{\dfrac{cosx}{sinx} - \dfrac{sinx}{cosx}} = \frac{1}{2}\cdot\sin4x (2)$

$\frac{cos^2{2x} - sin^2{2x} +1}{\dfrac{cos^2x-sin^2x}{sinx \cdot cosx}} =\frac{1}{2}\cdot\sin4x (3)$

now I don't now how to finish it

any help appreciated

2. In the third line you dropped 1 from the numerator. Use $\sin^2(2x)+\cos^2(2x)=1$ to substitute for 1. Also notice that $\cos^2(x)-\sin^2(x)=\cos(2x)$

That should keep you going.