Question

The absolute value of $\tan \frac{\pi }{16}+\tan \frac{5\pi }{16}+\tan \frac{9\pi }{16}+\tan \frac{13\pi }{16}$ is

Answer 1

Let $\theta =\frac{\pi }{16}\Rightarrow 8\theta =\frac{\pi }{2}$
$\tan \frac{\pi }{16}+\tan \frac{5\pi }{16}+\tan \frac{9\pi }{16}+\tan \frac{13\pi }{16}=\tan \theta +\tan 5\theta +\tan 9\theta +\tan 13\theta =\tan \theta +\tan 5\theta +\tan (8\theta +\theta )+\tan (8\theta +5\theta )=\tan \theta +\tan 5\theta +\tan (\frac{\pi }{2}+\theta )+\tan (\frac{\pi }{2}+5\theta )=\tan \theta +\tan 5\theta -\cot \theta -\cot 5\theta =\tan \theta -\cot \theta +\tan 5\theta -\cot 5\theta (\because \cot \theta -\tan \theta =2\cot 2\theta )=-2\cot 2\theta -2\cot 10\theta =-2[\cot 2\theta +\cot (8\theta +2\theta \left)\right]=-2[\cot 2\theta +\cot (\frac{\pi }{2}+2\theta )]=-2[\cot 2\theta -\tan 2\theta ]=-4\cot 4\theta =-4\cot \frac{\pi }{4}=-4$

Therefore, the absolute value of the expression is $4.$