Question

[(1+sin^2x)/(2+ctg^2x )]+[(1+cos^2x)/(2+tg^2x)]=1

se poate o rezolvare desfasurata ?

Answer 1

$\displaystyle \frac{1+\sin^2x}{2+\cot^2x}+\frac{1+\cos^2x}{2+\tan^2x}= \frac{1+\sin^2x}{2+\dfrac{\cos^2 x}{\sin^2 x}}+\frac{1+\cos^2x}{2+\dfrac{\sin^2 x}{\cos^2 x}} = \\ \\ \\ =\dfrac{1+\sin^2 x}{\dfrac{2\sin^2 x+\cos^2 x}{\sin^2 x}}+\dfrac{1+\cos^2 x}{\dfrac{2\cos^2 x+\sin^2 x}{\cos^2 x}} =$

$=\dfrac{\sin^2 x(1+\sin^2 x)}{2\sin^2+\cos^2 x}+\dfrac{\cos^2 x(1+\cos^2 x)}{2\cos^2 x+\sin^2 x}=\\ \\\\=\dfrac{\sin^2 x(1+\sin^2 x)}{\sin^2+\sin^2 x+\cos^2 x}+\dfrac{\cos^2 x(1+\cos^2 x)}{\cos^2 x+\cos^2 x+\sin^2 x}= \\ \\ \\=\dfrac{\sin^2 x(1+\sin^2 x)}{1+\sin^2 x}+\dfrac{\cos^2 x(1+\cos^2 x)}{1+\cos^2 x}=\\ \\ \\ = \sin^2 x+\cos^2 x = \boxed{1}$