Помогите пожалуйста!!! Докажите тождество:номера 9,10,11

Question 1

Question 2

Решите задачу:

$9)\; \; \frac{1+sina}{1+cosa}\cdot \frac{1+\frac{1}{cosa}}{1+\frac{1}{sina}}=\frac{1+sina}{1+cosa}\cdot \frac{\frac{cosa+1}{cosa}}{\frac{sina+1}{sina}}=\frac{1+sina}{1+cosa}\cdot \frac{(1+cosa)\cdot sina}{(1+sina)\cdot cosa}=\\\\=\frac{sina}{cosa}=tga\; \; ,\; \; tga=tga\; .$

$10)\; \; \frac{sin^4a-cos^4a+cos^2a}{2\, (1-cosa)}=\frac{(sin^2a-cos^2a)(sin^2a+cos^2a)+cos^2a}{2\, (1-cosa)}=\\\\=\frac{sin^2a-cos^2a+cos^2a}{2\cdot 2sin^2\frac{a}{2}}=\frac{sin^2a}{4\cdot sin^2\frac{a}{2}}=\frac{(2sin\frac{a}{2}\cdot cos\frac{a}{2})^2}{4sin^2\frac{a}{2}}=\frac{4sin^2\frac{a}{2}\cdot cos^2\frac{a}{2}}{4sin^2\frac{a}{2}}=\\\\=cos^2\frac{a}{2}=\frac{1+cosa}{2}=\frac{1}{2}\, (1+cosa)\; .$

$\frac{1}{2}\, (1+cosa)=\frac{1}{2}\, (1+cosa)\\\\\star \; \; sin^2a+cos^2a=1\; \; \star \\\\\star \; \; cos2a=cos^2a-sin^2a=cos^2a-(1-cos^2a)=2cos^2a-1\; \Rightarrow$

$cos^2a=\frac{1+cos2a}{2}\; \; \Rightarrow \; \; cos^2\frac{a}{2}=\frac{1+cosa}{2}\; \; \star \\\\\star sin^2a=\frac{1-cos2a}{2}\; \; \Rightarrow \; \; sin^2\frac{a}{2}=\frac{1-cosa}{2}\; \; \star \\\\\star \; \; sin2a=2\, sina\cdot cosa\; \; \Rightarrow \; \; sina=2\, sin\frac{a}{2}\cdot cos\frac{a}{2}\; \; \star \\\\11)\; \; \frac{1}{4\, sin^2a\cdot cos^2a}-\frac{(1-tg^2a)^2}{4\, tg^2a}=\frac{1}{(2\, sina\cdot cosa)^2}-\frac{\left(1-\frac{sin^2a}{cos^2a}\right )^2}{4tg^2a}=$

$=\frac{1}{sin^22a}-\frac{\frac{(cos^2a-sin^2a)^2}{cos^4a}}{4\cdot \frac{sin^2a}{cos^2a}}=\frac{1}{sin^22a}-\frac{\frac{cos^22a}{cos^4a}}{4\cdot \frac{sin^2a}{cos^2a}} =\frac{1}{sin^22a}-\frac{cos^22a\cdot cos^2a}{4\cdot sin^2a\cdot cos^4a}=\\\\=\frac{1}{sin^22a}-\frac{cos^22a}{4\, sin^2a\cdot cos^2a}=\frac{1}{sin^22a}-\frac{cos^22a}{sin^22a}=\frac{1-cos^22a}{sin^22a}=\frac{sin^22a}{sin^22a}=1\; ,\\\\1=1$