Question

Помогите решить ......................................

Answer 1

Ответ:

Объяснение:

1) $\frac{cos^3\beta-sin^3\beta }{1+cos\beta*sin\beta} =\frac{(cos\beta-sin\beta)(cos^2\beta+cos\beta*sin\beta+sin^2\beta)}{1+cos\beta*sin\beta}= \frac{(cos\beta-sin\beta)(1+cos\beta*sin\beta)}{1+cos\beta*sin\beta}=$

$= cos\beta-sin\beta$

2) $\frac{cos\beta }{1+sin\beta} - \frac{cos\beta }{1-sin\beta}=\frac{cos\beta(1-sin\beta)-cos\beta(1+sin\beta)}{(1+sin\beta)(1-sin\beta)} =\frac{cos\beta(1-sin\beta-1-sin\beta)}{1-sin^2\beta} =$

$\frac{cos\beta*(-2sin\beta)}{cos^2\beta} =\frac{-2sin\beta}{cos\beta} =-2tg\beta$

3) $(1+tg\beta)^2 + (1-tg\beta)^2 = 1 + 2tg\beta + tg^2\beta + 1 - 2tg\beta+tg^2\beta=2+2tg^2\beta=$

$=2(1+tg^2\beta)=\frac{2}{cos^2\beta}$

4) $\frac{1-4cos^2\beta*sin^2\beta}{(cos\beta+sin\beta)^2} +2cos\beta*sin\beta=\frac{1-(2cos\beta*sin\beta)^2}{cos^2\beta+2cos\beta*sin\beta+sin^2\beta} +2cos\beta*sin\beta=$

$\frac{(1-2cos\beta*sin\beta)(1+2cos\beta*sin\beta)}{1+2cos\beta*sin\beta} +2cos\beta*sin\beta=1-sin(2\beta)+sin(2\beta)=1$