Showing $\sqrt{\frac{1-\cos20^\circ}{1+\cos40^\circ}} =\frac{\cos80^\circ}{\cos20^\circ}$

Question

Here is a problem:

Here is (the short version of) the question:

I did this problem using the Cosine Rule on $\triangle PQU$ and $\triangle PTU$. This gave me an answer of $$\sqrt{\frac{1-\cos20^\circ}{1+\cos40^\circ}} \tag{1}$$ The given answer is, however: $$\frac{\cos80^\circ}{\cos20^\circ} \tag{2}$$ Both give the same numerical answer when put into a calculator.

I was wondering whether it is possible to convert/manipulate $(1)$ into $(2)$. Or, is using the Cosine Rule just not acceptable here, to get the "correct" answer?

Many thanks.

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If you don't understand how I got my answer or how they got theirs, I can attach both pieces of working. Hopefully it is clear though! Thanks again.

Answer 1

By the identities $1-\cos x=2\sin^2x/2$ and $1+\cos x=2\cos^2x/2$ your answer becomes $$\sqrt{\frac{2\sin^210^\circ}{2\cos^220^\circ}}=\frac{\sin10^\circ}{\cos20^\circ}=\frac{\cos80^\circ}{\cos20^\circ}$$