Question

12) tg(13pi-a)tg(13pi/2+a)-sin(a-22pi)

Answer 1

$\displaystile\bf\\12)\\\\tg\Big(13\pi -\alpha \Big)tg\Big(\dfrac{13\pi }{2} +\alpha \Big)-Sin\Big(\alpha -22\pi \Big)=\\\\\\=tg\Big(\pi -\alpha \Big)\cdot tg\Big(\dfrac{\pi }{2} +\alpha\Big)+Sin\alpha=\underbrace{-tg\alpha \cdot \Big(-Ctg\alpha \Big)}_{1}+Sin\alpha = Sin\alpha +1$