Question

ПОМАГИТЕ СРОЧНО, РЕШИТЕ

Answer 1

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Answer 2

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

$3)\; \; \frac{2sin^2a-1}{sina+cosa}=\frac{-cos2a}{\sqrt2\cdot sin(a+\frac{\pi}{4})}=-\frac{\sqrt2\, cos2a}{2\cdot sin(a+\frac{\pi}{4})}\\\\ili\; \; \frac{2sin^2a-1}{sina+cosa}=\frac{2sin^2a-cos^2a-sin^2a}{sina+cosa}=\frac{sin^2a-cos^2a}{sina+cosa}=\\\\=\frac{(sina-cosa)(sina+cosa)}{sina+cosa}=sina-cosa=\\\\=\sqrt2\cdot (\frac{1}{\sqrt2}\cdot sina-\frac{1}{\sqrt2}\cdot cosa)=\sqrt2\cdot sin(a-\frac{\pi }{4})\\\\\\\star \; \; tga=\frac{sina}{cosa}\; \; ,\; \; ctga=\frac{cosa}{sina} \; \; \star$

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