Question

!!!!SOOOS!!!!
30минут​

Answer 1

$\left ( 1+\mathrm{tg}^2\alpha +\frac{1}{\sin^2\alpha } \right )\cdot \sin^2\alpha \cos^2a=\left ( \frac{1}{\cos^2\alpha }+\frac{1}{\sin^2\alpha } \right )\cdot \sin^2\alpha \cos^2\alpha =$

$=\frac{\sin^2\alpha +\cos^2\alpha }{\sin^2\alpha \cos^2\alpha }\cdot \sin^2\alpha \cos^2\alpha =\frac{1}{ \sin^2\alpha \cos^2\alpha }\cdot \sin^2\alpha \cos^2\alpha =\frac{\sin^2\alpha \cos^2\alpha }{\sin^2\alpha \cos^2\alpha }=1$

$\frac{\sin\alpha +\cos \alpha }{\cos \left ( \pi/4-\alpha \right )}=\frac{\sqrt{2}\left ( 1/\sqrt{2}\sin \alpha +1/\sqrt{2}\cos \alpha \right )}{\cos\left ( \pi/4-\alpha \right )}=\frac{\sqrt{2}\left ( \cos \pi/4\sin \alpha +\sin \pi/4\cos \alpha \right )}{\cos\left ( \pi/4-\alpha \right )}=$$=\frac{\sqrt{2}\sin\left ( \alpha +\pi/4 \right )}{\cos\left ( \pi/4-\alpha \right )}=\frac{\sqrt{2}\cos\left ( \pi/2-\left ( \alpha +\pi/4 \right ) \right )}{\cos\left ( \pi/4-\alpha \right )}=\frac{\sqrt{2}\cos\left ( \pi/4-\alpha \right )}{\cos \left ( \pi/4-\alpha \right )}=\sqrt{2}$