Question

Solve

$\tan \frac{\pi }{16}+\tan \frac{3\pi }{16}+\tan \frac{5\pi }{16}+\tan \frac{7\pi }{16}$

Answer 1

$\Rightarrow \tan \frac{\pi }{16}+\tan \frac{3\pi }{16}+\tan \frac{5\pi }{16}+\tan \frac{7\pi }{16}$

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

$\Rightarrow \tan \frac{\pi }{16}+\tan \frac{3\pi }{16}+\cot \frac{\pi }{16}+\cot \frac{3\pi }{16}$

$\Rightarrow \frac{\sin \frac{\pi }{16}}{\cos \frac{\pi }{16}}+\frac{\cos \frac{\pi }{16}}{\sin \frac{\pi }{16}}+\frac{\sin \frac{3\pi }{16}}{\cos \frac{3\pi }{16}}+\frac{\cos \frac{3\pi }{16}}{\sin \frac{3\pi }{16}}$

$\Rightarrow \frac{{\sin }^2\frac{\pi }{16}+{\cos }^2\frac{\pi }{16}}{\sin \frac{\pi }{16}\cos \frac{\pi }{16}}+\frac{{\sin }^2\frac{3\pi }{16}+{\cos }^2\frac{3\pi }{16}}{\sin \frac{3\pi }{16}-\cos \frac{3\pi }{16}}$

$\because {\cos }^{2}x+{\sin }^{2}x=1$

$\Rightarrow \frac{1}{\sin \frac{\pi }{16}\cos \frac{\pi }{16}}+\frac{1}{\sin \frac{3\pi }{16}\cos \frac{3\pi }{16}}$

$\Rightarrow \frac{2}{2\sin \frac{\pi }{16}\cos \frac{\pi }{16}}+\frac{2}{2\sin \frac{3\pi }{16}\cos \frac{3\pi }{16}}$

$\because \sin 2x=2sinxcosx$

$\Rightarrow \frac{2}{\sin \frac{\pi }{8}}+\frac{2}{\sin \frac{3\pi }{8}}$

$\Rightarrow 2(\frac{1}{\sin \frac{\pi }{8}}+\frac{1}{\sin \frac{3\pi }{8}})$

$\Rightarrow 2\left(\frac{\sin \frac{3\pi }{8}+\sin \frac{\pi }{8}}{\sin \frac{\pi }{8}\sin \frac{3\pi }{8}}\right)$

$\because sinC+sinD=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

$\Rightarrow 2\left(\frac{2.\sin \left(\frac{\pi }{4}\right).\cos \left(\frac{\pi }{8}\right)}{\sin \frac{\pi }{8}.\sin \frac{3\pi }{8}}\right)\because \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$

$\Rightarrow 4\times \frac{1}{\sqrt{2}}\times \cot \frac{\pi }{8}cosec\left(\frac{3\pi }{8}\right)$

$\Rightarrow 2\sqrt{2}\cot \frac{\pi }{8}cosec\left(\frac{3\pi }{8}\right)$