Упростите выражение.......................

Question 1

Question 2

Решите задачу:

$\dfrac{1-2\sin^2 \alpha }{2\mathrm{ctg}( \frac{ \pi }{4}+ \alpha )\cdot\cos^2( \frac{ \pi }{4}- \alpha ) } = \dfrac{\cos2 \alpha }{2\mathrm{ctg}( \frac{ \pi }{4}+ \alpha )\cdot \frac{1+\cos2(\frac{ \pi }{4}- \alpha )}{2} } = \\\ =\dfrac{\cos2 \alpha }{\mathrm{ctg}( \frac{ \pi }{4}+ \alpha )\cdot (1+\cos(\frac{ \pi }{2}-2 \alpha ) ) } =\dfrac{\cos2 \alpha }{\mathrm{ctg}( \frac{ \pi }{4}+ \alpha )\cdot (1+\sin2 \alpha ) } =$
$=\dfrac{1 }{\mathrm{ctg}( \frac{ \pi }{4}+ \alpha) } \cdot \dfrac{\cos2 \alpha }{ 1+\sin2 \alpha } =\mathrm{tg}( \frac{ \pi }{4}+ \alpha )\cdot \dfrac{\cos^2 \alpha-\sin^2 \alpha }{ \cos^2 \alpha +\sin^2 \alpha +\sin2 \alpha } = \\\ = \dfrac{\mathrm{tg} \frac{ \pi }{4}+\mathrm{tg} \alpha }{1-\mathrm{tg} \frac{ \pi }{4} \mathrm{tg} \alpha } \cdot \dfrac{(\cos \alpha-\sin \alpha )(\cos \alpha+\sin \alpha )}{( \cos \alpha +\sin \alpha)^2 } =$
$= \dfrac{1+\mathrm{tg} \alpha }{1-\mathrm{tg} \alpha } \cdot \dfrac{\cos \alpha-\sin \alpha }{ \cos \alpha +\sin \alpha } = \dfrac{1+ \frac{\sin \alpha }{\cos \alpha } }{1-\frac{\sin \alpha }{\cos \alpha } } \cdot \dfrac{\cos \alpha-\sin \alpha }{ \cos \alpha +\sin \alpha } = \\\ = \dfrac{\cos \alpha + \sin \alpha }{\cos \alpha -\sin \alpha } \cdot \dfrac{\cos \alpha-\sin \alpha }{ \cos \alpha +\sin \alpha } =1$

Question 3

спасибо