Question

Aratati ca $\frac{1 + cos 2x}{1 - cos 2x} = ctg x$ pentru orice x ∈ ( 0 , $\frac{ \pi }{2}$ )

Answer 1

[tex]\frac{1+\cos2x}{1-\cos2x}=ctg^2 x\\ \\ \frac{1+\cos^2x-\sin^2x}{1-\cos^2x+\sin^2x}=ctg^2x\\ \\ \frac{\sin^2x+\cos^2x+\cos^2x-\sin^2x}{\sin^2x+\cos^2x- \\ \cos^2x+\sin^2x}=ctg^2x\\ \\ \frac{2\cdot \cos^2x}{2\cdot \sin^2x}=ctg^2x\\ \\ ctg^2x=ctg^2x(A)[/tex]