Question

упростить ((1+2〖sin〗^2 x)/(sin⁡x+cos⁡x ))-sin⁡x

Answer 1

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