Помогите пожалуйста!!!!!!!!!!!!!

Решите задачу:

$(sin^2t-tg^2t)\cdot \frac{ctg^2t}{sin^2t} = \frac{sin^2tctg^2t}{sin^2t} - \frac{tg^2tctg^2t}{sin^2t} = \frac{cos^2t-1}{sin^2t} = \frac{-sin^2t}{sin^2t} =-1$

$\frac{(tg\frac{\pi }{4}-ctg \frac{3\pi }{4})sin\frac{ \pi }{6}}{2cos\frac{ \pi }{3} -sin \frac{3 \pi }{4}} = \frac{(1-(-1))\frac{ 1 }{2}}{2\cdot\frac{ 1 }{2} - \frac{ \sqrt{2} }{2}} = \frac{2\cdot\frac{ 1 }{2}}{1- \frac{ \sqrt{2} }{2}} = \\\ = \frac{1}{1- \frac{ \sqrt{2} }{2}} = \frac{1+ \frac{ \sqrt{2} }{2}}{(1- \frac{ \sqrt{2} }{2})(1+ \frac{ \sqrt{2} }{2})} = \frac{1+ \frac{ \sqrt{2} }{2}}{ \frac{ 1 }{2}} =2+2 \sqrt{2}$