Тригонометрия ,помогите пожалуйста (хотя бы 2 и 4)

[latex] (\frac{1+\cos^2 \alpha}{\sin \alpha}-\sin \alpha)\cdot \frac{1}{2}tg \alpha= \frac{1+\cos^2\alpha-\sin^2 \alpha}{\sin \alpha}\cdot \frac{\sin \alpha}{2\cos \alpha}=\\\\= \frac{\cos^2\alpha+\cos^2\alpha}{\sin\alpha}\cdot \frac{\sin \alpha}{2\cos \alpha}= \frac{2\cos^2\alpha}{2\cos \alpha} =\cos \alpha[/latex] [latex]*******************************************[/latex] [latex] \frac{\sin^2\alpha}{\cos \alpha} - \frac{\cos^2 \alpha}{\sin \alpha} [/latex], если  [latex]\sin \alpha-\cos \alpha= \frac{1}{2} [/latex] [latex]\sin\alpha-\cos\alpha= \frac{1}{2}\\\\ (\sin\alpha-\cos\alpha)^2= \frac{1}{4}\\\\ \sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha= \frac{1}{4}\\\\ 1-2\sin\alpha\cos\alpha= \frac{1}{4}\\\\ -2\sin\alpha\cos\alpha=- \frac{3}{4} \\\\ \sin\alpha\cos\alpha= \frac{3}{4} \cdot \frac{1}{2}\\\\ \sin\alpha\cos\alpha = \frac{3}{8} [/latex] [latex]\frac{\sin^2\alpha}{\cos \alpha} - \frac{\cos^2 \alpha}{\sin \alpha}= \frac{(\sin \alpha-\cos\alpha)(\sin^2 \alpha+\sin\alpha\cdot \cos\alpha+\cos^2 \alpha)}{\sin\alpha\cdot\cos\alpha} =\\\\ \frac{ \frac{1}{2}(\sin^2 \alpha+\sin\alpha\cdot \cos\alpha+\cos^2 \alpha) }{\sin\alpha\cdot\cos\alpha}= \frac{ \frac{1}{2}(1+ \frac{3}{8}) }{ \frac{3}{8} }= \frac{1}{2}\cdot \frac{11}{8} \cdot \frac{8}{3}= \frac{11}{6} [/latex]