Решите пожалуйста )))))))))))))))

$1.\;\frac{\sin^2\alpha-1}{1-\cos^2\alpha}=\frac{\sin^2\alpha-\sin^2\alpha-\cos^2\alpha}{\sin^2\alpha+\cos^2\alpha-\cos^2\alpha}=\frac{-\cos^2\alpha}{\sin^2\alpha}=-tg^2\alpha\\\alpha=\frac\pi4\Rightarrow -tg^2\alpha=-(1)^2=-1\\\\2.\;\cos^2\alpha+ctg^2\alpha+\sin^2\alpha=1+ctg^2\alpha=1+\frac{\cos^2\alpha}{\sin^2\alpha}=\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}=\\=\frac1{\sin^2\alpha}\\\alpha=\frac\pi6\Rightarrow\frac1{\sin^2\alpha}=\frac1{\left(\frac12\right)^2}=\frac1{\frac14}=4$

$3.\;\frac1{\cos^2\alpha}-1=\frac{1-\cos^2\alpha}{\cos^2\alpha}=\frac{\sin^2\alpha+\cos^2\alpha-\cos^2\alpha}{\cos^2\alpha}=\frac{\sin^2\alpha}{\cos^2\alpha}=tg^2\alpha\\\alpha=\frac\pi3\Rightarrow tg^2\alpha=(\sqrt3)^2=3\\\\4.\;\cos^2\alpha+tg^2\alpha ctg^2\alpha+\sin^2\alpha=1+\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\frac{\cos^2\alpha}{\sin^2\alpha}=1+1=2$
Последнее равенство справедливо для любых α