Answer to Question #273152 in Trigonometry for seiji

Question

Find the solutions of the equation if 0≤𝑡<2𝜋

tan^2 𝑡−1=0

Answer 1

$\tan^2t-1=0$

$(\tan t-1)(\tan t+1)=0$

$\tan t=\pm1$

$t=\dfrac{\pi}{4}+\dfrac{\pi n}{2}, n\in \Z$

If we consider $0\leq t<2\pi$

$t_1=\dfrac{\pi}{4}, t_2=\dfrac{3\pi}{4}, t_3=\dfrac{5\pi}{4}, t_4=\dfrac{7\pi}{4}$

$\{\dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}\}$