Question

((tg2a + 1/cos2a).(cosa - sina))/sqrt2cos(pi/4-a)

Answer 1

$\star \; \; tg2a+\frac{1}{cos2a}=\frac{sin2a}{cos2a}+\frac{1}{cos2a}=\frac{sin2a+cos2a}{cos2a}\; ;\\\\\star \; \; cosa+sina=\sqrt2\cdot (\frac{1}{\sqrt2}\cdot cosa+\frac{1}{\sqrt2}\cdot sina)=\\=\sqrt2(sin\frac{\pi }{4}\cdot cosa+cos\frac{\pi}{4}\cdot sina)=\sqrt2\cdot sin(\frac{\pi}{4}+a)\; ;\\\\\star \; \; sin2a+cos2a=\sqrt2(\frac{1}{\sqrt2}\cdot sin2a+\frac{1}{\sqrt2}\cdot cos2a)=\\=\sqrt2(cos\frac{\pi}{4}\cdot sin2a+sin\frac{\pi }{4}\cdot cos2a)=\sqrt2\cdot sin(\frac{\pi }{4}+2a)\; ;\\\\\star \; \; cos(\frac{\pi}{4}-a)=sin(\frac{\pi}{2}-(\frac{\pi}{4}-a))=sin(\frac{\pi}{4}+a)\; ;\\\\\star \; \; cos2a=cos^2a-sin^2a\; ;$

$\frac{(tg2a+\frac{1}{cos2a})\cdot (cosa-sina)}{\sqrt2\cdot cos(\frac{\pi}{4}-a)}=\frac{\frac{sin2a+cos2a}{cos2a}\cdot (cosa-sina)}{\sqrt2\cdot cos(\frac{\pi}{4}-a)}=\\\\=\frac{\sqrt2\cdot sin(\frac{\pi}{4}+2a)\cdot (cosa-sina)}{(cos^2a-sin^2a)\cdot \sqrt2\, cos(\frac{\pi}{4}-a)}=\frac{sin(\frac{\pi}{4}+2a)}{(cosa+sina)\cdot cos(\frac{\pi}{4}-a)}=\\\\=\frac{sin(\frac{\pi}{4}+2a)}{\sqrt2\cdot sin(\frac{\pi}{4}+a)\cdot sin(\frac{\pi}{4}+a)}=\frac{sin(\frac{\pi}{4}+2a)}{\sqrt2\cdot sin^2(\frac{\pi}{4}+a)}$