Question

3) cos(2pi - alpha) * (tg((3pi)/2 - alpha)) ^ 2 * tg(2pi + alpha) * (ctg(pi/2 + alpha)) ^ 2

Answer 1

$\displaystyle\bf\\Cos\Big(2\pi -\alpha \Big)\cdot\Bigg[tg\Big(\frac{3\pi }{2} -\alpha \Big)\Bigg]^{2} \cdot tg\Big(2\pi +\alpha \Big)\cdot\Bigg[Ctg\Big(\frac{\pi }{2} +\alpha \Big)\Bigg]^{2} =\\\\\\=Cos\alpha \cdot Ctg^{2} \alpha\cdot tg\alpha \cdot tg^{2} \alpha =\\\\\\=Cos\alpha \cdot\frac{Sin\alpha }{Cos\alpha } \cdot\underbrace{\Big(Ctg^{2} \alpha \cdot tg^{2} \alpha \Big)}_{1}=Sin\alpha\\\\\\Otvet \ : \ Sin\alpha$