Find the value of

$(1+\cot A-\mathrm{cosec}A)(1+\tan A+\sec A)-3({\sin }^{2}A+{\cos }^{2}A).$

The correct answer is -1.

To find the value of the given trigonometric expression, we can break it down into step-by-step simplifications using standard trigonometric identities.

The given expression is:

$(1+\cot A-\mathrm{cosec}A)(1+\tan A+\sec A)-3({\sin }^{2}A+{\cos }^{2}A)$

Step 1: Convert all trigonometric terms in the first two factors to sine and cosine.
Recall the quotient and reciprocal identities:
$\cot A=\frac{\cos A}{\sin A}$
$\mathrm{cosec}A=\frac{1}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\sec A=\frac{1}{\cos A}$

Substitute these into the first factor:
$1+\cot A-\mathrm{cosec}A=1+\frac{\cos A}{\sin A}-\frac{1}{\sin A}=\frac{\sin A+\cos A-1}{\sin A}$

Substitute these into the second factor:
$1+\tan A+\sec A=1+\frac{\sin A}{\cos A}+\frac{1}{\cos A}=\frac{\cos A+\sin A+1}{\cos A}=\frac{\sin A+\cos A+1}{\cos A}$

Step 2: Multiply the simplified factors.
$(1+\cot A-\mathrm{cosec}A)(1+\tan A+\sec A)=\frac{(\sin A+\cos A-1)(\sin A+\cos A+1)}{\sin A\cos A}$

Notice that the numerator is of the difference of squares form $(x-1)(x+1)=x^2-1$, where $x=\sin A+\cos A$.
$(\sin A+\cos A-1)(\sin A+\cos A+1)={(\sin A+\cos A)}^2-1^2$

Expand ${(\sin A+\cos A)}^2$:
${(\sin A+\cos A)}^2={\sin }^{2}A+{\cos }^{2}A+2\sin A\cos A$

Using the Pythagorean identity ${\sin }^{2}A+{\cos }^{2}A=1$:
${(\sin A+\cos A)}^2=1+2\sin A\cos A$

Substitute this back into the numerator:
${(\sin A+\cos A)}^2-1=1+2\sin A\cos A-1=2\sin A\cos A$

Therefore, the product simplifies to:
$\frac{2\sin A\cos A}{\sin A\cos A}=2$

Step 3: Simplify the second half of the given expression.
The second part is $-3({\sin }^{2}A+{\cos }^{2}A)$.
Since ${\sin }^{2}A+{\cos }^{2}A=1$, this becomes:
$-3\left(1\right)=-3$

Step 4: Combine the results.
Subtracting the second part from the first part gives:
$2-3=-1$

Thus, the final value of the expression is -1.