Sin(a-b)/sin(a+b) и cos(a-b)/cos(a+b) Выразите через а)tga и tgb б)ctga и ctgb

Question 1

Question 2

$\frac{sin( \alpha - \beta )}{sin( \alpha + \beta } = \frac{sin \alpha *cos \beta -cos \beta *sin \alpha }{sin \alpha *cos \beta +cos \alpha *sin \beta } = \frac{ \frac{sin \alpha *cos \beta }{cos \alpha *cos \beta }- \frac{cos \ \alpha *sin \ \beta }{cos \alpha *cos \beta } }{ \frac{sin \alpha *cos \beta }{cos \alpha *cos \beta } + \frac{cos \alpha *sin \beta }{cos \alpha *cos \beta } } = \frac{tg \alpha -tg \beta }{tg \alpha +tg \beta }$

2) $\frac{Cos( \alpha - \beta )}{cos( \alpha + \beta )} = \frac{cos \alpha *cos \beta -sin \alpha *sin \beta }{cos \alpha *cos \beta +sin \alpha *sin \beta } = \frac{ \frac{cos \alpha *cos \beta }{cos \alpha *cos \beta } - \frac{sin \alpha *sin \beta }{cos \alpha *cos \beta } }{ \frac{cos \alpha *cos \beta }{cos \alpha *cos \beta }+ \frac{sin \alpha *sin \beta }{cos \alpha *cos \beta } } = \frac{1-tg \alpha *tg \beta }{1+tg \alpha *tg \beta }$

3) $\frac{sin( \alpha - \beta )}{sin( \alpha + \beta )} = \frac{sin \alpha *cos \beta -cos \alpha *sin \beta }{sin \alpha *cos \beta +sin \beta *cos \alpha } = \frac{ \frac{sin \alpha *cos \beta }{sin \alpha *sin \beta }- \frac{cos \alpha *sin \beta }{sin \alpha *sin \beta } }{ \frac{sin \alpha *cos \beta }{sin \alpha *sin \beta }+ \frac{cos \alpha *sin \beta }{sin \alpha *sin \beta } } = \frac{ctg \beta -ctg \alpha }{ctg \beta +ctg \alpha }$

4) $\frac{cos( \alpha - \beta )}{cos( \alpha + \beta )} = \frac{cos \alpha *cos \beta +sin \alpha sin \beta }{cos \alpha *cos \beta -sin \alpha *sin \beta } = \frac{ \frac{cos \alpha *cos \beta }{sin \alpha *sin \beta }+ \frac{sin \alpha *sin \beta }{sin \alpha *sin \beta } }{ \frac{cos \alpha *cos \beta }{sin \alpha *sin \beta } - \frac{sin \alpha *sin \beta }{sin \alpha *sin \beta } } = \frac{ctg \alpha *ctg \beta +1}{ctg \alpha *ctg \beta -1}$

Question 3

во втором задании после знака равно в числителе и знаменателе знаки поменяйте на противоположные везде