Answer
The expression in terms of the function $\sec x\ \text{ and }\ \tan x$ is $\sec x\tan x$.
Work Step by Step
By using the reciprocal identity of trigonometry $cscx=\frac{1}{sinx}$.
And the above expression can be further solved as,
$\begin{align}
& \frac{1}{\csc x-\sin x}=\frac{1}{\frac{1}{\sin x}-\sin x} \\
& =\frac{1}{\frac{1-{{\sin }^{2}}x}{\sin x}}
\end{align}$
Apply the Pythagorean identity of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
Now, the expression ${{\cos }^{2}}x$ can be written as $1-{{\sin }^{2}}x$. so,
$\frac{1}{\frac{1-{{\sin }^{2}}x}{\sin x}}=\frac{1}{\frac{{{\cos }^{2}}x}{\sin x}}$
Then, rewrite the main fraction bar as
$\frac{1}{\frac{{{\cos }^{2}}x}{\sin x}}=1\div \frac{{{\cos }^{2}}x}{\sin x}$
Invert the divisor and multiply; then, we get:
$1\div \frac{{{\cos }^{2}}x}{\sin x}=1\times \frac{\sin x}{{{\cos }^{2}}x}$
And the above expression is simplified as,
$1\times \frac{\sin x}{{{\cos }^{2}}x}=\frac{1}{\cos x}\times \frac{\sin x}{\cos x}$
Now, apply the reciprocal identity of trigonometry $\sec x=\frac{1}{\cos x}$ , and the quotient identity of trigonometry $\tan x=\frac{\sin x}{\cos x}$.
And the above expression can be further solved as,
$\begin{align}
& \frac{1}{\cos x}\times \frac{\sin x}{\cos x}=\sec x\times \tan x \\
& =\sec x\tan x
\end{align}$
Thus, the given expression is written in the terms of the function $\sec x\ \text{ and }\ \tan x$ as $\sec x\tan x$.
Hence, the expression $\frac{1}{\csc x-\sin x}$ in terms of the function $\sec x\ \text{ and }\ \tan x$ is $\sec x\tan x$.
