Question

помогите с алгеброй, объясните как делать 774

Answer 1

$1)\; \; ctg\beta -\frac{cos\beta -1}{sin\beta}=\frac{cos\beta }{sin\beta }-\frac{cos\beta -1}{sin\beta }=\frac{cos\beta -cos\beta +1}{sin\beta }=\frac{1}{sin\beta }=cosec\beta \\\\2)\; \; \frac{sin^2-1}{cos^2a-1}+tga\cdot ctga=\frac{-cos^2a}{-sin^2a}+1=\frac{cos^2a+sin^2a}{sin^2a}=\frac{1}{sin^2a}\; \; \; \Rightarrow \\\\\star \; \; ctg^2a+1=\frac{1}{sin^2a}\; \; \star \\\\3)\; \; \frac{1}{sina-1}-\frac{1}{sina+1}=\frac{sina+1-(sina-1)}{(sina-1)(sina+1)}=\frac{2}{sin^2a-1}=-\frac{2}{cos^2a}$

$4)\; \; tg^2\beta (sin^2\beta -1)=\frac{sin^2\beta }{cos^2\beta } \cdot (-cos^2\beta )=-sin^2\beta \\\\5)\; \; \frac{1-ctg\gamma}{tg\gamma-1}=\frac{1-\frac{cos\gamma }{sin\gamma }}{\frac{sin\gamma}{cos\gamma}-1}=\frac{(sin\gamma-cos\gamma)\cdot cos\gamma}{sin\gamma \cdot (sin\gamma-cos\gamma)}=\frac{cos\gamma }{sin\gamma }=ctg\gamma \\\\6)\; \; cos^2a-(ctg^2a+1)sin^2a=cos^2a-\frac{1}{sin^2a}\cdot sin^2a=cos^2a-1=-sin^2a$