$\begin{align}  & \text{Let }y=\frac{1+\sin x+\cos x}{1-\sin x+\cos x} \\ & \text{Using quotient rule } \\ & \frac{dy}{dx}=\frac{(1-\sin x+\cos x)\tfrac{d}{dx}(1+\sin x+\cos x)-(1+\sin x+\cos x)\tfrac{d}{dx}(1-\sin x+\cos x)}{{{(1-\sin x+\cos x)}^{2}}} \\ & \frac{dy}{dx}=\frac{(1-\sin x+\cos x)(\cos x-\sin x)-(1+\sin x+\cos x)(-\cos x-\sin x)}{{{(1-\sin x+\cos x)}^{2}}} \\ & \frac{dy}{dx}=\frac{(1-\sin x+\cos x)(\cos x-\sin x)+(1+\sin x+\cos x)(\cos x+\sin x)}{{{(1-\sin x+\cos x)}^{2}}} \\ & \frac{dy}{dx}=\frac{\left[ \cos x-\sin x\cos x+{{\cos }^{2}}x-\sin x+{{\sin }^{2}}x-\sin x\cos x \right]+\left[ \cos x+\sin x\cos x+{{\cos }^{2}}x+\sin x+{{\sin }^{2}}x+\sin x\cos x \right]}{{{(1-\sin x+\cos x)}^{2}}} \\ & \frac{dy}{dx}=\frac{2\cos x+2{{\sin }^{2}}x+2{{\cos }^{2}}x}{{{(1-\sin x+\cos x)}^{2}}}=\frac{2\cos x+2}{{{(1-\sin x+\cos x)}^{2}}} \\ & \frac{dy}{dx}=\frac{2(\cos x+1)}{1-2\sin x\cos x-2\sin x+2\cos x+{{\sin }^{2}}x+{{\cos }^{2}}x} \\ & \frac{dy}{dx}=\frac{2(\cos x+1)}{1-2\sin x\cos x-2\sin x+2\cos x+1}=\frac{2(\cos x+1)}{2-2\sin x\cos x-2\sin x+2\cos x} \\ & \frac{dy}{dx}=\frac{2(\cos x+1)}{2(1-\sin x\cos x-\sin x+\cos x)}=\frac{\cos x+1}{1+\cos x-\sin x\cos x-\sin x} \\ & \frac{dy}{dx}=\frac{\cos x+1}{(1+\cos x)-\sin x(\cos x+1)}=\frac{\cos x+1}{(1-\sin x)(\cos x+1)} \\ & \frac{dy}{dx}=\frac{1}{1-\sin x} \\\end{align}$