Question

Помогите пожалуйста решить !!!​

Answer 1

$1)\frac{tg\frac{\alpha}{4}Cos\frac{\alpha}{4}}{1+Ctg^{2}\frac{\alpha}{4}}=\frac{\frac{Sin\frac{\alpha}{4}}{Cos\frac{\alpha}{4}}*Cos\frac{\alpha}{4}}{\frac{1}{Sin^{2}\frac{\alpha }{4}}} =Sin\frac{\alpha}{4}*Sin^{2}\frac{\alpha}{4}=Sin^{3}\frac{\alpha}{4}$

$5)(tg3\beta +Ctg3\beta )*Sin^{2}3\beta =(\frac{Sin3\beta}{Cos3\beta}+\frac{Cos3\beta}{Sin3\beta})*Sin^{2}3\beta =\frac{Sin^{2}3\beta+Cos^{2}3\beta}{Sin3\beta Cos3\beta}*Sin^{2}3\beta =\frac{1}{Sin3\beta Cos3\beta}*Sin^{2}3\beta=\frac{Sin3\beta}{Cos3\beta}=tg3\beta$

$6)\frac{Cos^{2}4\alpha}{1+tg^{2}4\alpha(Sin^{2}4\alpha-1)}=\frac{Cos^{2}4\alpha}{1+tg^{2}4\alpha*(-Cos^{2}4\alpha)}= \frac{Cos^{2}4\alpha}{1-\frac{Sin^{2}4\alpha }{Cos^{2}4\alpha}*Cos^{2}4\alpha}=\frac{Cos^{2}4\alpha}{1-Sin^{2}4\alpha}=\frac{Cos^{2}4\alpha}{Cos^{2}4\alpha} =1$