Решите плез все 3 сижу ломаю мозг вообще не вкуриваю

Решите задачу:

$6)\quad sinx= \frac{3}{5} \\\\\frac{\pi}{2}\ \textless \ x\ \textless \ \pi \; \; \to \; \; cosx\ \textless \ 0\; ,\; \; tgx\ \textless \ 0\\\\cosx=-\sqrt{1-sin^2x}=-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}\\\\tgx=\frac{sinx}{cosx}=\frac{3/5}{-4/5}=-\frac{3}{4}\\\\cosx-tgx=-\frac{4}{5}-(-\frac{3}{4})=-\frac{4}{5}+\frac{3}{4}=-\frac{1}{20}$

$7)\quad \frac{tg(\pi -a)}{cos(\pi +a)} \cdot \frac{sin(\frac{3\pi}{2}+a)}{tg(\frac{3\pi}{2}+a)} = \frac{-tga}{-cosa} \cdot \frac{-cosa}{-ctga} =-tg^2a\\\\8)\quad \alpha - \beta =\frac{\pi}{2}\; \; \to \; \; \alpha =\frac{\pi}{2}+ \beta \\\\\frac{sin \alpha -sin \beta }{cos \alpha +cos \beta }= \frac{2sin\frac{ \alpha - \beta }{2}cos\frac{ \alpha + \beta }{2}}{2cos\frac{ \alpha - \beta }{2}cos\frac{ \alpha + \beta }{2}} = \frac{sin\frac{\alpha -\beta }{2}}{cos\frac{\alpha -\beta }{2}} = \frac{sin\frac{\pi}{4}}{cos\frac{\pi}{4}} =tg\frac{\pi}{4}=1$