Question

Упростите:

$\frac{1+cosx}{sinx} - \frac{sinx}{1-cosx}$

Answer 1

$\frac{1+cosx}{sinx} - \frac{sinx}{1-cosx} = \frac{(1+cosx)(1-cosx) - sin(x)^2}{sinx(1-cosx)} = \frac{1-cos(x)^2 - sin(x)^2}{sinx(1-cosx)} =$
$\frac{sin(x)^2 - sin(x)^2}{sinx(1-cosx)} = \frac{0}{sinx(1-cosx)} = 0$