[Euler-Fermat]

$tan\frac{7\pi}{16} + tan\frac{3\pi}{8} = cosec\frac{\pi}{8}$

$tan\frac{7\pi}{16} + tan\frac{3\pi}{8} = \dfrac{sin\frac{7\pi}{16}cos\frac{3\pi}{8}+sin\frac{3\pi}{8}cos\frac{7\pi}{16}}{cos\frac{7\pi}{16}cos\frac{3\pi}{8}} = \dfrac{sin\frac{13\pi}{16}}{cos\frac{7\pi}{16}cos\frac{3\pi}{8}} $

$cos\frac{7\pi}{16}cos\frac{3\pi}{8} = 2cos\frac{\pi}{16}[sin\frac{\pi}{16}]^2
= sin\frac{\pi}{8}sin\frac{\pi}{16}$

$\therefore \dfrac{sin\frac{13\pi}{16}}{cos\frac{7\pi}{16}cos\frac{3\pi}{8}} = \dfrac{sin\frac{13\pi}{16}}{sin\frac{\pi}{8}sin\frac{\pi}{16}} \not=
cosec\frac{\pi}{8}$