Question

Помогите решить!
$1) \frac{cosx}{1-sinx} = 1+sinx\ 2)\frac{cosx}{sinx+1} = sinx-1\ 3)\frac{sinx}{2cos^{2}x\frac{x}{2} } -2sin^{2}\frac{x}{2}$

Answer 1

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

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

$3)\frac{sin^{2} x}{2cos^{2}\frac{x}{2} } = 2sin^{2}\frac{x}{2} \\ \frac{sin ^{2} x \times 2 {sin}^{2} \frac{x}{2} }{2cos^{2}\frac{x}{2} \times 2 {sin}^{2} \frac{x}{2} } = 2sin^{2}\frac{x}{2} \\ \frac{sin ^{2} x \times 2 {sin}^{2} \frac{x}{2} }{sin^{2}x }= 2sin^{2}\frac{x}{2} \\2sin^{2}\frac{x}{2}= 2sin^{2}\frac{x}{2}$