Precalculus (6th Edition) Blitzer

The expression in terms of the function $sinx$ is $\frac{1}{sinx}$.
By using the reciprocal identity of trigonometry $\sec x=\frac{1}{\cos x}$ , and $cscx=\frac{1}{sinx}$ , and the quotient identity of trigonometry $\tan x=\frac{\sin x}{\cos x}$. The above expression can be further solve as, $\frac{\sec x+\csc x}{1+\tan x}=\left( \frac{\frac{1}{\cos x}+\frac{1}{\sin x}}{1+\frac{\sin x}{\cos x}} \right)$ Now, multiply the numerator and denominator of $\frac{\sec x+\csc x}{1+\tan x}$ by $\sin x\cos x$; then we get, $\frac{\sec x+\csc x}{1+\tan x}=\left( \frac{\frac{1}{\cos x}+\frac{1}{\sin x}}{1+\frac{\sin x}{\cos x}} \right)\frac{\sin x\cos x}{\sin x\cos x}$ Further solve as, \begin{align} & \left( \frac{\frac{1}{\cos x}+\frac{1}{\sin x}}{1+\frac{\sin x}{\cos x}} \right)\frac{\sin x\cos x}{\sin x\cos x}=\frac{\frac{\sin x\cos x}{\cos x}+\frac{\sin x\cos x}{\sin x}}{\sin x\cos x+\frac{{{\sin }^{2}}x\cos x}{\cos x}} \\ & =\frac{\frac{{{\sin }^{2}}x\cos x+\sin x{{\cos }^{2}}x}{\cos x\sin x}}{\frac{\sin x{{\cos }^{2}}x+{{\sin }^{2}}x\cos x}{\cos x}} \end{align} Now, invert the above expression and multiply. Then we get, \begin{align} & \frac{\frac{{{\sin }^{2}}x\cos x+\sin x{{\cos }^{2}}x}{\cos x\sin x}}{\frac{\sin x{{\cos }^{2}}x+{{\sin }^{2}}x\cos x}{\cos x}}=\frac{{{\sin }^{2}}x\cos x+\sin x{{\cos }^{2}}x}{\cos x\sin x}\times \frac{\cos x}{\sin x{{\cos }^{2}}x+{{\sin }^{2}}x\cos x} \\ & =\frac{1}{\sin x} \end{align} Thus, the provided expression is written in the terms of the function $\sin x$ as $\frac{1}{sinx}$. Hence, the expression $\frac{\sec x+\csc x}{1+\tan x}$ in terms of the function $sinx$ is $\frac{1}{sinx}$.