Question

Evaluate: ${\tan }^2\frac{\pi }{16}+{\tan }^2\frac{2\pi }{16}+{\tan }^2\frac{3\pi }{16}+\dots \dots +{\tan }^2\frac{7\pi }{16}$

Answer 1

$\frac{7\pi }{16}=\frac{\pi }{2}-\frac{\pi }{16}$

$\frac{6\pi }{16}=\frac{\pi }{2}-\frac{2\pi }{16}$

$\frac{5\pi }{16}=\frac{\pi }{2}-\frac{2\pi }{16}$

$\frac{4\pi }{16}=\frac{\pi }{4}$

$\Rightarrow {\tan }^2\frac{\pi }{16}+{\tan }^2\frac{2\pi }{16}+{\tan }^2\frac{3\pi }{16}+...\dots ..{\tan }^2\frac{6\pi }{16}+{\tan }^2\frac{7\pi }{16}$

$\Rightarrow {\tan }^2\frac{\pi }{16}+{\tan }^2\frac{2\pi }{16}+{\tan }^2\frac{3\pi }{16}+...\dots +{\tan }^2(\frac{\pi }{3}-\frac{2\pi }{16})+{\tan }^2(\frac{\pi }{2}-\frac{\pi }{16})$

$\Rightarrow {\tan }^2\frac{\pi }{16}+{\tan }^2\frac{2\pi }{16}{\tan }^2\frac{3\pi }{16}+...\dots ..+{\cot }^2\frac{2\pi }{16}+{\cot }^2\frac{\pi }{16}$

$\Rightarrow ({\tan }^2\frac{\pi }{16}+{\cot }^2\frac{\pi }{16})+({\tan }^2\frac{2\pi }{16}+{\cot }^2\frac{2\pi }{16})+({\tan }^2\frac{3\pi }{16}+{\cot }^2\frac{3\pi }{16})+{\tan }^2\frac{4\pi }{16}$

Solve first bracket

$({\tan }^2\frac{\pi }{16}+{\cot }^2\frac{\pi }{16})={(\tan \frac{\pi }{16}+\cot \frac{\pi }{16})}^2-2\tan \frac{\pi }{16}\cot \frac{\pi }{16}$

$={(\frac{\sin \pi /16}{\cos \pi /16}+\frac{\cos \pi /16}{\sin \pi /16})}^2-2$

$={\left(\frac{t\times 2}{2\sin \left(\pi /16\right)\cos \left(\pi /16\right)}\right)}^2-2$

$={\left(\frac{2}{\sin \pi /8}\right)}^2-2$

$=\frac{4\times 2}{2{\sin }^2\pi /8}-2$

$=\frac{8}{(1-\cos \pi /4)}-2$

$=\frac{8}{(1-\frac{1}{\sqrt{2}})}-2$

$=\frac{8\sqrt{2}}{(\sqrt{2}-1)}-2$

$\Rightarrow 8\sqrt{2}(\sqrt{2}+1)-2$

Similarly $2^{nd}$ bracket$=\frac{8}{(1-\cos \frac{\pi }{2})}-2=8-2=6$

& similarly $3^{rd}$ bracket$=\frac{8}{(1-\cos 3\pi /4)}-2=8\sqrt{2}(\sqrt{2}-1)-2$

Thus, equation $\left(1\right)$

$\Rightarrow 8\sqrt{2}(\sqrt{2}+1)-2+6+1+8\sqrt{2}(\sqrt{2}-1-2=$

$\Rightarrow 16+8\sqrt{2}-2+7+8\sqrt{2}\times \sqrt{2}-8\sqrt{2}-2=$

$\Rightarrow 35$.