If \(y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{\cos x}}}\), then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) is given by

\(\begin{aligned} & y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+\frac{\cos x}{\ldots \ldots}}}} \\ & y=\frac{\sin x}{1+\frac{\cos x}{1+y}}\end{aligned}\)
\(\begin{aligned}
& y=\frac{(1+y) \sin x}{(1+y)+\cos x} \\
& \dot{y}((1+y)+\cos x)=(1+y) \sin x \\
& y+y^2+y \cos x=\sin x+\sin x \cdot y
\end{aligned}\)
Differentiating w.r.to \(x\), we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}+y(-\sin x)+\cos x \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}\)
\(=\frac{\mathrm{d} y}{\mathrm{~d} x} \sin x+(1+y) \cos x\)
\(\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}-y \sin x+\cos x \frac{\mathrm{~d} y}{\mathrm{~d} x} \\ & =\sin x \frac{\mathrm{~d} y}{\mathrm{~d} x}+(1+y) \cos x \\ & \frac{\mathrm{~d} y}{\mathrm{~d} x}(1+2 y+\cos x-\sin x)\end{aligned}\)
\(=\cos x+y \cos x+y \sin x\)
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y \sin x+(1+y) \cos x}{1+2 y+\cos x-\sin x}\)