Question

помогите пожалуйста срочно надо пожалуйста​

Answer 1

$a)\ \ \dfrac{\pi}{2}<a<\pi \ \ \Rightarrow \ \ sina>0\ ,\ cosa<0\\\\\sqrt{1-cos^2a}-\sqrt{1-sin^2a}=\sqrt{sin^2a}-\sqrt{cos^2a}=|\underbrace {sina}_{>0}|-|\underbrace{cosa}_{<0}|=\\\\=sina-(-cosa)=sina+cosa=sina+sin\Big(\dfrac{\pi}{2}-a\Big)=\\\\=2\cdot sin\dfrac{\pi}{4}\cdot cos\Big(a-\dfrac{\pi}{4}\Big)=\sqrt2\cdot \dfrac{\sqrt2}{2}\cdot cos\Big(a-\dfrac{\pi}{4}\Big)=\sqrt2\cdot cos\Big(a-\dfrac{\pi}{4}\Big)$

$b)\ \ \dfrac{3\pi }{2}<a<2\pi \ \ \ \Rightarrow \ \ \ sina<0\ \ ,\ \ cosa>0\\\\sqrt{1+tg^2a}-\sqrt{1+ctg^2a}=\sqrt{\dfrac{1}{cos^2a}}-\sqrt{\dfrac{1}{sin^2a}}=\dfrac{1}{|cosa|}-\dfrac{1}{|sina|}=\\\\\=\dfrac{1}{cosa}-\dfrac{1}{-sina}=\dfrac{1}{cosa}+\dfrac{1}{sina}=\dfrac{sina+cosa}{sina\cdot cosa}=$

$=\dfrac{1}{cosa}-\dfrac{1}{-sina}=\dfrac{1}{cosa}+\dfrac{1}{sina}=\dfrac{sina+cosa}{sina\cdot cosa}=\dfrac{2\cdot (sina+cosa)}{sin2a}=\\\\\\=\dfrac{2\sqrt2\cdot cos(a-\frac{\pi}{4})}{sin2a}$