Rezolvați logtanA+logtanB=0

$$\log \tan A + \log \tan B = 0$$

Rezolvați pentru A

$$A=\pi n_{3}+\frac{\pi -2B}{2}$$

$$n_{3}\in \mathrm{Z}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(B>-\pi n_{1}+\pi n_{3}-\pi \text{ and }B<\pi n_{3}-\pi n_{1}\right)$$

$$\exists n_{2}\in \mathrm{Z}\text{ : }\left(B>\pi n_{2}\text{ and }B<\pi n_{2}+\frac{\pi }{2}\right)$$

Rezolvați pentru B

$$B=\pi n_{3}+\frac{\pi -2A}{2}$$

$$n_{3}\in \mathrm{Z}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(A>-\pi n_{1}+\pi n_{3}-\pi \text{ and }A<\pi n_{3}-\pi n_{1}\right)$$

$$\exists n_{2}\in \mathrm{Z}\text{ : }\left(A>\pi n_{2}\text{ and }A<\pi n_{2}+\frac{\pi }{2}\right)$$