Answer
The expression in terms of the function $tanx$ is $\tan x$.
Work Step by Step
By using the function given to express the trigonometric relationship between the expression and the function:
$\left( \sec x+\csc x \right)\left( \sin x+\cos x \right)-2-\cot x=\sec x\sin x+\sec x\cos x+\csc x\sin x+\csc x\cos x-2-\cot x$
By using the reciprocal identity of trigonometry $\sec x=\frac{1}{\cos x}$ , and $cscx=\frac{1}{sinx}$.
And the above expression can be further solve as,
$\frac{1}{\cos x}\times \sin x+\frac{1}{\cos x}\times \cos x+\frac{1}{\sin x}\times \sin x+\frac{1}{\sin x}\times \cos x-2-\cot x=\frac{\sin x}{\cos x}+1+1+\frac{\cos x}{\sin x}-2-\cot x$
Apply the quotient identity of trigonometry $tanx=\frac{sinx}{\cos x}$ , and $\cot x=\frac{\cos x}{sinx}$.
And the above expression can be further solve as,
$\begin{align}
& \frac{\sin x}{\cos x}+1+1+\frac{\cos x}{\sin x}-2-\cot x=\tan x+2+\cot x-2-\cot x \\
& =\tan x
\end{align}$
Thus, the given expression is written in terms of the function $\tan x$ as $\tan x$.
Hence, the expression $\left( \sec x+\csc x \right)\left( \sin x+\cos x \right)-2-\cot x$ in terms of the function $\tan x$ is $\tan x$.
