Solving problem with basic trigonometry formula

 Prove that $ \displaystyle \frac{2}{{1+\sin x+\cos x}}=\frac{{\sin x}}{{1+\cos x}}+\frac{{\cos x}}{{1+\sin x}}$.


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

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

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

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

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{1-\cos x}}{{\sin x}}+\frac{{1-\sin x}}{{\cos x}}$

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

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

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin x}}{{1+\cos x}}+\frac{{\cos x}}{{1+\sin x}}$