Question

Упростите выражение sin ² (3 пи /2 - a) + cos²(3 пи /2 - а) / ctg² (пи - а) * tg²( пи - а)

Answer 1

Решение:

$\frac{sin^{2}(\frac{3\pi}{2}-\alpha)+cos^{2}(\frac{3\pi}{2}-\alpha)}{ctg^{2}(\pi-\alpha)*tg^{2}(\pi-\alpha)}=\frac{1}{(-ctg\alpha)^{2}*(-tg\alpha)^{2}}=\frac{1}{ctg^{2}\alpha*tg^{2}\alpha}=\frac{1}{(ctg\alpha*tg\alpha)^{2}}=\frac{1}{(\frac{1}{tg\alpha}*tg\alpha)^{2}}=\frac{1}{1^{2}}=\frac{1}{1}=1$

Нужные формулы:

$sin^{2}(\frac{3\pi}{2}-\alpha)+cos^{2}(\frac{3\pi}{2}-\alpha) = 1$

$ctg^{2}(\pi-\alpha)=(ctg(\pi-\alpha))^{2}=(-ctg\alpha)^{2}=ctg^{2}\alpha$

$tg^{2}(\pi-\alpha)=(tg(\pi-\alpha))^{2} = (tg\alpha)^{2} = tg^{2}\alpha$

$ctg\alpha=\frac{cos\alpha}{sin\alpha}=\frac{1}{tg\alpha}$

Ответ: 1

Answer 2

$\frac{Sin^{2}(\frac{3\pi }{2}-\alpha)+Cos^{2}(\frac{3\pi }{2} -\alpha)}{Ctg^{2}(\pi-\alpha)\cdot tg^{2}(\pi-\alpha)}=\frac{Cos^{2}\alpha+Sin^{2}\alpha}{Ctg^{2}\alpha\cdot tg^{2} \alpha}=\frac{1}{1} =\boxed1$