Question

Compute the value of the expression , $\frac{1}{5}[{\tan }^2\frac{\pi }{16}+{\tan }^2\frac{2\pi }{16}+{\tan }^2\frac{3\pi }{13}+...+{\tan }^2\frac{7\pi }{16}]$.

Answer 1

$\tan \theta +\cot \theta =\sec \theta \csc \theta$

${\tan }^2\theta +{\cot }^2\theta +2={\sec }^2\theta {\csc }^2\theta$

${\tan }^2\theta +{\cot }^2\theta =\frac{4}{{\sin }^2\left(2\theta \right)}-2$

${\tan }^2\frac{\pi }{16}+{\tan }^2\frac{4\pi }{16}={\tan }^2\left(\frac{\pi }{16}\right)+{\cot }^2\left(\frac{\pi }{16}\right)$

$=\frac{4}{{\sin }^2\left(\frac{\pi }{8}\right)}-2$

${\tan }^2\frac{2\pi }{16}+{\tan }^2\frac{6\pi }{16}={\tan }^2\left(\frac{2\pi }{16}\right)+{\cot }^2\left(\frac{2\pi }{16}\right)$

$=\frac{4}{{\sin }^2\left(\frac{\pi }{4}\right)}-2=6$

${\tan }^2\frac{3\pi }{16}+{\tan }^2\frac{5\pi }{16}={\tan }^2\left(\frac{3\pi }{16}\right)+{\cot }^2\left(\frac{3\pi }{16}\right)$

$=\frac{4}{{\sin }^2\left(\frac{3\pi }{8}\right)}-2=\frac{4}{{\cos }^2\frac{\pi }{8}}-2$

$\frac{1}{5}\left[{\tan }^2\right(\frac{\pi }{16})+{\tan }^2(\frac{2\pi }{16})+.........+{\tan }^2(\frac{7\pi }{16}\left)\right]$

$=\frac{1}{5}\left[\right(\frac{4}{{\sin }^2\left(\frac{\pi }{8}\right)}-2)+6+(\frac{4}{{\cos }^2\left(\frac{\pi }{8}\right)}-2)+{\tan }^2(\frac{4\pi }{16}\left)\right]$

$=\frac{1}{5}\left[4\right(\frac{1}{{\sin }^2\left(\frac{\pi }{8}\right)}+\frac{1}{{\cos }^2\left(\frac{\pi }{8}\right)})+3]$

$=\frac{1}{5}[\frac{4}{{\sin }^2\left(\frac{\pi }{8}\right){\cos }^2\left(\frac{\pi }{8}\right)}+3]$

$=\frac{1}{5}[\frac{16}{{\sin }^2\left(\frac{\pi }{4}\right)}+3]$

$=\frac{1}{5}(32+3)=7$