Question

Помогите решить: (1/tgx) + (1/ctgx) - 2ctg2x = 2

Answer 1

$\frac{1}{tgx}+\frac{1}{ctgx}-2ctg2x=ctgx+tgx-2ctg2x=\\\\=\frac{cosx}{sinx}+\frac{sinx}{cosx}-\frac{2cos2x}{sin2x}=\frac{cos^2x+sin^2x}{sinx\cdot cosx}-\frac{2cos2x}{sin2x}=\\\\=\frac{1}{\frac{1}{2}sin2x}-\frac{2cos2x}{sin2x}=\frac{2}{sin2x}-\frac{2\, cos2x}{sin2x}=\frac{2\, (1-cos2x)}{sin2x}=\\\\= \frac{2\cdot 2\, sin^2x}{2\, sinx\cdot cosx}=\frac{2\, sinx}{cosx}=2tgx\\\\2tgx\ne 2\; \; \; (!)\\\\\star \; \; sin^2x=\frac{1-cos2x}{2}\; \; \to \; \; \; 1-cos2x=2sin^2x$