Respuesta :
[tex]\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}[/tex]
therefore, that fraction will become undefined if the denominator ever turns to 0.
now, recall that cos(π/2) is 0, and cos(3π/2) is 0, and cos(5π/2) is also zero and so on, thus
[tex]\bf tan\left( \frac{\pi }{2} \right)=\cfrac{sin\left( \frac{\pi }{2} \right)}{cos\left( \frac{\pi }{2} \right)}\implies tan\left( \frac{\pi }{2} \right)=\cfrac{sin\left( \frac{\pi }{2} \right)}{0} \\\\\\ tan\left( \frac{3\pi }{2} \right)=\cfrac{sin\left( \frac{3\pi }{2} \right)}{cos\left( \frac{3\pi }{2} \right)}\implies tan\left( \frac{3\pi }{2} \right)=\cfrac{sin\left( \frac{3\pi }{2} \right)}{0} [/tex]
[tex]\bf tan\left( \frac{5\pi }{2} \right)=\cfrac{sin\left( \frac{5\pi }{2} \right)}{cos\left( \frac{5\pi }{2} \right)}\implies tan\left( \frac{5\pi }{2} \right)=\cfrac{sin\left( \frac{5\pi }{2} \right)}{0} \\\\\\ tan\left( \frac{7\pi }{2} \right)=\cfrac{sin\left( \frac{7\pi }{2} \right)}{cos\left( \frac{7\pi }{2} \right)}\implies tan\left( \frac{7\pi }{2} \right)=\cfrac{sin\left( \frac{7\pi }{2} \right)}{0}[/tex]
[tex]\bf tan\left( \frac{\pi }{2}+\pi \right)=\cfrac{sin\left( \frac{\pi }{2}+\pi \right)}{cos\left( \frac{\pi }{2}+\pi \right)}\implies tan\left( \frac{\pi }{2}+\pi \right)=\cfrac{sin\left( \frac{\pi }{2}+\pi \right)}{0} \\\\\\ domain\implies \{x|x\in \mathbb{R}, ~~x\ne \frac{\pi }{2}+n\pi,~~n\in \mathbb{Z} \}[/tex]
therefore, that fraction will become undefined if the denominator ever turns to 0.
now, recall that cos(π/2) is 0, and cos(3π/2) is 0, and cos(5π/2) is also zero and so on, thus
[tex]\bf tan\left( \frac{\pi }{2} \right)=\cfrac{sin\left( \frac{\pi }{2} \right)}{cos\left( \frac{\pi }{2} \right)}\implies tan\left( \frac{\pi }{2} \right)=\cfrac{sin\left( \frac{\pi }{2} \right)}{0} \\\\\\ tan\left( \frac{3\pi }{2} \right)=\cfrac{sin\left( \frac{3\pi }{2} \right)}{cos\left( \frac{3\pi }{2} \right)}\implies tan\left( \frac{3\pi }{2} \right)=\cfrac{sin\left( \frac{3\pi }{2} \right)}{0} [/tex]
[tex]\bf tan\left( \frac{5\pi }{2} \right)=\cfrac{sin\left( \frac{5\pi }{2} \right)}{cos\left( \frac{5\pi }{2} \right)}\implies tan\left( \frac{5\pi }{2} \right)=\cfrac{sin\left( \frac{5\pi }{2} \right)}{0} \\\\\\ tan\left( \frac{7\pi }{2} \right)=\cfrac{sin\left( \frac{7\pi }{2} \right)}{cos\left( \frac{7\pi }{2} \right)}\implies tan\left( \frac{7\pi }{2} \right)=\cfrac{sin\left( \frac{7\pi }{2} \right)}{0}[/tex]
[tex]\bf tan\left( \frac{\pi }{2}+\pi \right)=\cfrac{sin\left( \frac{\pi }{2}+\pi \right)}{cos\left( \frac{\pi }{2}+\pi \right)}\implies tan\left( \frac{\pi }{2}+\pi \right)=\cfrac{sin\left( \frac{\pi }{2}+\pi \right)}{0} \\\\\\ domain\implies \{x|x\in \mathbb{R}, ~~x\ne \frac{\pi }{2}+n\pi,~~n\in \mathbb{Z} \}[/tex]