[tex]tg(x+\pi)=\frac{\sin(x+\pi)}{\cos(x+\pi)}=\frac{\sin x\cos\pi+\cos x\sin\pi}{\cos x\cos\pi-\sin x\sin\pi}=\frac{-\sin x}{-\cos x}=tg\:x=2\\\text{Sau te po\c ti folosi de faptul c\u a tangenta este func\c tie periodic\u a,}\\\text{de perioad\u a }T=\pi\text{. Prin urmare, }tg(x+k\pi)=tg\:x,\:k\in\mathbb{Z},\:\forall\:x\in\mathbb{R}.\\\text{Deci, }tg(x+\pi)=tg\:x=2.[/tex]
