Question

помогите пожалуйста
1) вычислить:
$36 \sqrt{3 \times \tan \binom{\pi}{3} } \times \sin \binom{\pi}{6}$
$54 \sqrt{3} \times \tan \binom{\pi}{6} \times \sin \binom{\pi}{6}$
2) упростить:
${ \cos }^{2} (2\pi - t) + { \cos }^{2} ( \frac{3\pi}{2} + t)$
$\frac{1 - \cos {}^{2} \alpha }{ \sin^{2} \alpha }$
$\sin^{2} (\pi + t) + \sin ^{2} ( \frac{\pi}{2} + t)$
$\frac{1 - { \sin }^{2} \alpha }{ \cos^{2} \alpha }$

​

Answer 1

$1)36\sqrt{3}*tg\frac{\pi }{3}*Sin\frac{\pi }{6}=36\sqrt{3}*\sqrt{3}*\frac{1}{2}=36*3*\frac{1}{2}=54\\\\2)54\sqrt{3}*tg\frac{\pi }{6}*Sin\frac{\pi }{6}=54\sqrt{3}*\frac{1}{\sqrt{3}}*\frac{1}{2}=54*\frac{1}{2}=27$

$3)Cos^{2}(2\pi-t)+Cos^{2}(\frac{3\pi }{2}+t)=Cos^{2}t+Sin^{2}t=1\\\\4)\frac{1-Cos^{2}\alpha}{Sin^{2}\alpha} =\frac{Sin^{2}\alpha}{Sin^{2}\alpha}=1$

$5)Sin^{2}(\pi+t)+Sin^{2}(\frac{\pi }{2}+t)=Sin^{2}t+Cos^{2}t=1\\\\6)\frac{1-Sin^{2}\alpha}{Cos^{2}\alpha}=\frac{Cos^{2}\alpha}{Cos^{2} \alpha}=1$