Question

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Answer 1

$1)\; \; sina=\frac{4}{5}\; \; ,\; \; cosa=\pm \sqrt{1-sin^2a}\\\\0<a<\frac{\pi }{2}\; \; \to \; \; cos>0\; \; \to \; \; cosa=+\sqrt{1-\frac{16}{25}}=\frac{3}{5}\\\\cos\beta =-\frac{5}{13}\; \; \to \; \; sina=\pm \sqrt{1-cos^2a}\\\\\pi <\beta <\frac{3\pi }{2}\; \; \to \; \; sin\beta <0\; \; \to \; \; sin\beta =-\sqrt{1-\frac{25}{169}}=-\frac{12}{13}\\\\cos(a+\beta )=cosa\, cos\beta-sina\. sin\beta =-\frac{3}{5}\cdot \frac{5}{13}+\frac{4}{5}\cdot \frac{12}{13}=\frac{33}{65}$

$2)\; \; \frac{tg(\frac{5\pi }{4}-a)(1+sin2a)}{cos(\frac{5\pi }{2}-2a)}=\frac{tg(\pi +\frac{\pi}{4}-a)(1+sin2a)}{cos(2\pi +\frac{\pi}{2}-2a)}=\frac{tg(\frac{\pi}{4}-a)(1+sin2a)}{sin2a}=\\\\=tg(\frac{\pi }{4}-a)\cdot \frac{1+sin2a}{sin2a}=\frac{1-tga}{1+tga}\cdot \frac{sin^2a+cos^2a+2sina\, cosa}{sin2a}=\\\\=\frac{1-\frac{sina}{cosa}}{1+\frac{sina}{cosa}}\cdot \frac{(sina+cosa)^2}{sin2a} =\frac{cosa-sina}{cosa+sina}\cdot \frac{(sina+cosa)^2}{sin2a}=\frac{(cosa-sina)(sina+cosa)}{sin2a}=\\\\=\frac{cos^2a-sin^2a}{sin2a}=\frac{cos2a}{sin2a}=tg2a$