Answer
The expression in terms of the function $sinx$ is $\frac{1}{sinx}$.
Work Step by Step
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 $\frac{1}{sinx}$.
