Prove that $$\sec(\pi/7)\frac{(4-\sec^2(\pi/7))}{(2-\sec^2(\pi/7))}=4$$
MY ATTEMPT: I thought of writing $\sec^2(\pi/3)$ in place of 4 and $\sec^2(\pi/4)$ in place of 2 in the question, and then applying $\cos^2(A)-\cos^2(B)=\sin(A+B)\sin(B-A)$, but it didn't helped (it made the question more complicated).
I also tried converting the whole question to $\tan$, but it didn't helped too.
Can anyone help me out with a solution. Thanks.
Best Answer
$$ \begin{align} \sec(\pi/7)\frac{(4-\sec^2(\pi/7))}{(2-\sec^2(\pi/7))} &= \frac{1}{\cos(\pi/7)}\cdot\frac{4\cos^2(\pi/7)-1}{2\cos^2(\pi/7)-1}\\ &= \frac{2\cos(2\pi/7)+1}{\cos(\pi/7)\cos(2\pi/7)}\\ &= \frac{2\sin(\pi/7)(2\cos(2\pi/7)+1)}{(2\sin(\pi/7)\cos(\pi/7))\cos(2\pi/7)}\\ &= 4\frac{\sin(\pi/7)(2\cos(2\pi/7)+1)}{\sin(4\pi/7)}\\ &= 4\frac{(\cos(2\pi/7)\sin(\pi/7)+2\sin(\pi/7)\cos^2(\pi/7))}{\sin(4\pi/7)}\\ &= 4\frac{(\cos(2\pi/7)\sin(\pi/7)+\sin(2\pi/7)\cos(\pi/7))}{\sin(4\pi/7)}\\ &= 4\frac{\sin(3\pi/7)}{\sin(4\pi/7)}\\ & = 4 \end{align} $$