Prove LHS=RHS for the following expression
\[\dfrac{{\cot A - 1}}{{\cot A + 1}} = \dfrac{{\cos A - \sin A}}{{\cos A + \sin A}}\] \[\]

Hint: In order to verify the expression we will take the complex side of the equation, simplify it using reciprocal and quotient identities. The nature of the identities will depend on the nature of the question. After successful application the identities we will simplify the question until we get the same expression as on the other side.

Formula Used: In the given question we have used a very basic yet simple trigonometric value that is,\[\dfrac{{\cos A}}{{\sin A}} = cotA\]. There are multiple ways of writing various trigonometric functions

Complete step-by-step solution:
Given,
 \[\Rightarrow \dfrac{{\cot A - 1}}{{\cot A + 1}} = \dfrac{{\cos A - \sin A}}{{\cos A + \sin A}}\]
We have
\[\Rightarrow RHS = \dfrac{{\cos A - \sin A}}{{\cos A + \sin A}}\]
Dividing the numerator and denominator by\[\sin A\], we get
\[\Rightarrow \dfrac{{\dfrac{{\cos A - \sin A}}{{\sin A}}}}{{\dfrac{{\cos A + \sin A}}{{\sin A}}}}\]
Further we simplify the equation to get,
\[\Rightarrow \dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}}}}\]
We know that the reciprocal identity of cotangent function is
\[\Rightarrow \dfrac{{\cos A}}{{\sin A}} = cotA\]
Thus, we substitute this value in equation
We get,
\[ \Rightarrow \dfrac{{cotA + 1}}{{cotA - 1}} = LHS\]
Hence. Proved.

Note: In general, identities help in solving trigonometric equations. The most important identities are the reciprocal identities and quotient identities.
They are as follows,
Reciprocal Identities for sine and cosecant function
\[\sin A = \dfrac{1}{{\cos ecA}}, \cos ecA = \dfrac{1}{{\sin A}}\],
Reciprocal Identities for cosine and secant function
\[\cos A = \dfrac{1}{{\sec A}}, \sec A = \dfrac{1}{{\cos A}}\],
Reciprocal Identities for tangent and cotangent function
\[\tan A = \dfrac{1}{{\cot A}}\]And \[\cot A = \dfrac{1}{{\tan A}}\]
The quotient identities define relationships among various trigonometric functions, they are follows:
\[\tan A = \dfrac{{\sin A}}{{\cos A}}\]
\[cotA = \dfrac{{\cos A}}{{\sin A}}\]

To prove trigonometric equations we should usually start with the complicated side of the question and keep simplifying the equation until it has been transformed into the same expression as on the other side. The other methods of solving a trigonometric equation involve expanding the expressions, factoring the expression or simply using basic algebraic strategies to obtain desired results. Simplifying one side of the equation equal to the other side is one of the most commonly used methods.