How to simplify the expression $\csc18°*\sec36°$?

Question

Please help me, I tried several methods but I do not understand how I can solve this other than substituting the actual values in the expression.

Answer 1

We have $$\csc18 \sec36=\frac{1}{\sin 18\cos 36}=\frac{2\cos 18}{\sin 36\cos 36}=\frac{4\cos 18}{\sin 72}=\frac{4\sin 72}{\sin 72}=4,$$ where I have used the transformation $2\sin x\cos x=\sin (2x)$ repeatedly, and $\cos(90-x)=\sin x$ in the penultimate equality. Finally, all arguments are in degrees.

Answer 2

$$\sin 18^o = \frac{\sqrt5 -1}{4} \text{ and } \cos 36^o = \frac{\sqrt5 +1}{4}$$ If you didn't know these you can search their proofs on the internet. Thus, $$\csc 18^o = \frac{4}{\sqrt5 -1} \text{ and } \sec 36^o = \frac{4}{\sqrt5 +1}$$ $$\implies \csc 18^o \cdot \sec 36^o = \frac{4}{\sqrt5 -1}\cdot\frac{4}{\sqrt5 +1}= \frac{16}{4} = 4$$

Answer 3

Using the same knowledge as in @rash's answer, an alternative is to let $\theta=\frac{\pi}{10}$ so you want $$\frac{1}{\sin\theta\cos 2\theta}=\frac{2}{\sin 3\theta-\sin\theta}=\frac{2}{\frac{\sqrt{5}+1}{4}-\frac{\sqrt{5}-1}{4}}=4.$$