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**Hint:**Here, we will use the Trigonometric Identity, trigonometric ratios and a suitable Algebraic Identity to prove the given trigonometric function. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right angled triangle with respect to any of its acute angle.

**Formula Used:**

We will use the following formula:

1. Trigonometric Identity: \[{\sin ^2}x + {\cos ^2}x = 1\].

2. The difference between the square of the numbers is given by an Algebraic Identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].

3. Trigonometric Ratio: \[\csc A = \dfrac{1}{{\sin A}}\] and \[\cot A = \dfrac{{\cos A}}{{\sin A}}\].

**Complete Step by Step Solution:**

We are given that the trigonometric function \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \csc A + \cot A\].

Now, we will multiply the numerator and the denominator of the given fraction by \[\sin A\], so we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\left( {\cos A - \sin A + 1} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Now, by multiplying the terms with the terms inside the Parentheses only in the numerator, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\cos A - {{\sin }^2}A + \sin A}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Trigonometric Identity: \[{\sin ^2}x + {\cos ^2}x = 1\]

Now, by rewriting the equation using the Trigonometric Identity, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\cos A + \sin A - \left( {1 - {{\cos }^2}A} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

The difference between the square of the numbers is given by an Algebraic Identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Now, by using an Algebraic Identity, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\cos A + \sin A - \left( {1 + \cos A} \right)\left( {1 - \cos A} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Now, by taking out the common terms, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\sin A\left( {1 + \cos A} \right) - \left( {1 + \cos A} \right)\left( {1 - \cos A} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Again by taking out the common terms, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\left( {1 + \cos A} \right)\left( {\sin A - 1 + \cos A} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\left( {1 + \cos A} \right)\left( {\cos A + \sin A - 1} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}\]

Now, by cancelling out the common terms, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\left( {1 + \cos A} \right)}}{{\sin A}}\]

Now, by segregating the terms, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{1}{{\sin A}} + \dfrac{{\cos A}}{{\sin A}}\]

Trigonometric Ratio: \[\csc A = \dfrac{1}{{\sin A}}\] and \[\cot A = \dfrac{{\cos A}}{{\sin A}}\]

Now, by using the Trigonometric Ratio, we get

\[ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \csc A + \cot A\]

**Therefore, \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \csc A + \cot A\]is proved.**

**Note:**

We know that Trigonometric Equation is defined as an equation involving trigonometric ratios. Trigonometric identity is an equation which is always true for all the variables. We should know that we have many trigonometric identities which are related to all the other trigonometric equations. Trigonometric Ratios are used to find the relationships between the sides of a right angle triangle.

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