Question

помогите пожалуйста ​

Answer 1

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

найдем значения выражения если sinx=0,4

$2 \times {0.4}^{2} - 1 = 2 \times0.16 - 1 = 0.32 - 1 = - 0.68$

..........

$\frac{2 \cos( \alpha + \beta ) +4sin \alpha sin \beta }{ \sin( \alpha + \beta ) - 2 \cos \alpha \sin \beta } = \\ = \frac{2cos \alpha cos \beta - 2sin \alpha sin \beta + 4sin \alpha sin \beta }{cos \alpha sin \beta + sin \alpha cos \beta - 2cos \alpha sin \beta } = \\ = \frac{2cos \alpha cos \beta + 2sin \alpha sin \beta }{sin \alpha sin \beta - cos \alpha sin \beta } = \\ = \frac{2 \cos( \alpha - \beta ) }{ \sin( \alpha - \beta ) } = 2ctg (\alpha - \beta )$

$\alpha - \beta = 120 ^ \circ$

$2ctg (\alpha - \beta ) = 2ctg {120}^ \circ = - \frac{ 2\sqrt{3} }{3}$