Question

помогите с 248 ; 250 ! Спасибо заранее​

Answer 1

$248.\; \; \; \; cos(45^\circ -a)=\frac{1}{3}\; \; \Rightarrow \\\\cos45^\circ \cdot cosa+sin45^\circ \cdot sina=\frac{1}{3}\\\\\frac{\sqrt2}{2}\cdot cosa+\frac{\sqrt2}{2}\cdot sina=\frac{1}{3}\; \; \to \; \; \; \frac{\sqrt2}{2}\cdot (cosa+sina)=\frac{1}{3}\; \to \\\\sina+cosa=\frac{\sqrt2}{3}\\\\(sina+cosa)^2=\frac{2}{9}\\\\\underbrace {sin^2a+cos^2a}_{1}+\underbrace {2\, sina\cdot cosa}_{sin2a}=\frac{2}{9}\\\\1+sin2a=\frac{2}{9}\\\\\frac{1+sin2a}{sina+cosa}=\frac{2}{9}:\frac{\sqrt2}{3}=\frac{2\cdot 3}{9\cdot \sqrt2}=\frac{\sqrt2}{3}$

$250.\; \; \; \frac{1}{sin170^\circ }-\frac{\sqrt3}{sin100^\circ }=\frac{1}{sin(180^\circ -10^\circ )}-\frac{\sqrt3}{sin(180^\circ -80^\circ )}=\\\\=\frac{1}{sin10^\circ }-\frac{\sqrt3}{sin80^\circ }=\frac{sin80^\circ -\sqrt3\cdot sin10^\circ }{sin10^\circ \cdot sin80^\circ }=\frac{2\cdot (\frac{1}{2}\cdot sin80^\circ -\frac{\sqrt3}{2}\cdot sin10^\circ )}{sin10^\circ \cdot sin80^\circ }=\\\\=\frac{2\cdot (\cos60^\circ \cdot sin(90^\circ -10^\circ )-sin60^\circ \cdot sin10^\circ )}{\frac{1}{2}\cdot (cos70^\circ -cos90^\circ )}=$

$=\frac{2\cdot (cos60^\circ \cdot cos10^\circ -sin60^\circ \cdot sin10^\circ )}{\frac{1}{2}\cdot (cos70^\circ -0)}=\frac{2\cdot cos70^\circ }{\frac{1}{2}\cdot cos70^\circ }=4$