Помогите 1 , 2 задания

Решите задачу:

$1.\;a)\;\frac{\sin2a}{\sin^2a}=\frac{2\sin a\cos a}{\sin^2a}=\frac{2\cos a}{\sin a}=2ctg\;a\\b)\;\frac{\sin2a}{2\sin a}=\frac{2\sin a\cos a}{2\sin a}=\cos a\\c)\;\sin^2a+\cos2a=\sin^2a+\cos^2a-\sin^2a=\cos^2a\\d)\;\frac{\cos2a}{\cos a-\sin a}-\sin a=\frac{\cos^2a-\sin^2a}{\cos a-\sin a}-\sin a=\\=\frac{(\cos a-\sin a)(\cos a+\sin a)}{\cos a-\sin a}-\sin a=\cos a+\sin a-\sin a=\cos a$

$2.\;\sin a=\frac5{13}\\\sin^2a+\cos^2a=1\\\cos a=\sqrt{1-\sin^2a}=\sqrt{1-\left(\frac5{13}\right)^2}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\pm\frac{12}{13}\\\frac\pi2<a<\pi\Rightarrow\cos a=-\frac{12}{13}\\\sin2a=2\sin a\cos a=2\cdot\frac5{13}\cdot(-\frac{12}{13})=-\frac{120}{169}\\\cos2a=\cos^2a-\sin^2a=\left(-\frac{12}{13}\right)^2-\left(\frac{15}{13}\right)^2=\frac{144}{169}-\frac{25}{169}=\frac{119}{169}$