Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Let the given expression be $E$.$$E=\frac{\cos (\frac{\pi }{14})\cos (\frac{3\pi }{14})\cos (\frac{5\pi }{14})}{\cos (\frac{2\pi }{14})\cos (\frac{4\pi }{14})\cos (\frac{6\pi }{14})}$$ Let $x=\frac{\pi }{14}$. Then $7x=\frac{\pi }{2}$. The expression becomes $E=\frac{\cos x\cos 3x\cos 5x}{\cos 2x\cos 4x\cos 6x}$. Using the complementary angle identity $\cos \theta =\sin (\frac{\pi }{2}-\theta )$, we can rewrite the terms in the numerator:$\cos x=\cos (\frac{\pi }{14})=\sin (\frac{\pi }{2}-\frac{\pi }{14})=\sin (\frac{6\pi }{14})=\sin 6x$.$\cos 3x=\cos (\frac{3\pi }{14})=\sin (\frac{\pi }{2}-\frac{3\pi }{14})=\sin (\frac{4\pi }{14})=\sin 4x$.$\cos 5x=\cos (\frac{5\pi }{14})=\sin (\frac{\pi }{2}-\frac{5\pi }{14})=\sin (\frac{2\pi }{14})=\sin 2x$. Substitute these into the numerator of $E$: $$E=\frac{\sin 6x\sin 4x\sin 2x}{\cos 2x\cos 4x\cos 6x}=\left(\frac{\sin 2x}{\cos 2x}\right)\left(\frac{\sin 4x}{\cos 4x}\right)\left(\frac{\sin 6x}{\cos 6x}\right)$$ $$E=\tan 2x\tan 4x\tan 6x=\tan \left(\frac{2\pi }{14}\right)\tan \left(\frac{4\pi }{14}\right)\tan \left(\frac{6\pi }{14}\right)$$ $$E=\tan \left(\frac{\pi }{7}\right)\tan \left(\frac{2\pi }{7}\right)\tan \left(\frac{3\pi }{7}\right)$$ For an odd integer $n$, the product ${\prod }_{k=1}^{(n-1)/2}\tan \left(\frac{k\pi }{n}\right)=\sqrt{n}$. Here, $n=7$, so $(n-1)/2=3$. Thus, $E=\tan \left(\frac{\pi }{7}\right)\tan \left(\frac{2\pi }{7}\right)\tan \left(\frac{3\pi }{7}\right)=\sqrt{7}$.