СРОЧНО ПОМОГИТЕЕЕ :с 1)найдите sin a - cos a, если tg a = -(3/4) и pi/2 < a < pi...

$1)\; tg \alpha =-\frac{3}{4},\; \frac{\pi}{2}< \alpha <\pi \\\\1+tg^2 \alpha =\frac{1}{cos^2 \alpha }\; \; \to \; \; cos^2 \alpha =\frac{1}{1+tg^2 \alpha }=\frac{1}{1+\frac{9}{16}}=\frac{16}{25}\\\\ \alpha \in 2\; chetverti\; \to \; cos \alpha <0,\; cos \alpha =-\frac{4}{5}\\\\1+ctg^2 \alpha =\frac{1}{sin^2 \alpha }\; \; \to \; sin^2 \alpha =\frac{1}{1+ctg^2 \alpha }=\frac{1}{1+\frac{1}{tg^2 \alpha }}=\frac{1}{1+\frac{16}{9}}=\frac{9}{25}\\\\ \alpha \in 2\; chetverti\; \to \; sin \alpha =+\frac{3}{5}$

0,\; cos \alpha =+\frac{1}{\sqrt5}" alt="sin \alpha -cos \alpha =\frac{3}{5}-(-\frac{4}{5})=\frac{7}{5}\\\\2)\; tg \alpha =2,\; 0< \alpha <\frac{\pi}{2}\\\\cos^2 \alpha =\frac{1}{1+tg^2 \alpha }=\frac{1}{1+4}=\frac{1}{5},cos\alpha=\pm\frac{1}{\sqrt5}\\\\ \alpha \in 1 \; chetverti\; \to \; cos \alpha >0,\; cos \alpha =+\frac{1}{\sqrt5}" align="absmiddle" class="latex-formula">