Question

Prove that \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A\]

Answer 1

Hint:
Here, we will divide the left hand side of the expression by \[\sin A\] and simplify it further to get the expression in the form of cosecant and cotangent function. We will then use suitable trigonometric and algebraic identity to simplify the expression further and prove that the given fraction in the LHS is equal to the sum of trigonometric functions in the RHS.
Formula Used:
We will use the following formulas:
\[1 = \cos e{c^2}A - {\cot ^2}A\]
\[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\]

Complete step by step solution:
We will first take into consideration the left hand side of the equation.
LHS \[ = \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}}\]
Now, dividing both the numerator as well as denominator by \[\sin A\], we get,
\[ \Rightarrow \] LHS \[ = \dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}}}}\]
Using \[\dfrac{{\cos A}}{{\sin A}} = \cot A\] and \[\cos ecA = \dfrac{1}{{\sin A}}\], we get
\[ \Rightarrow \] LHS \[ = \dfrac{{\cot A - 1 + \cos ecA}}{{\cot A + 1 - \cos ecA}}\]
Now substituting \[1 = \cos e{c^2}A - {\cot ^2}A\] in the numerator, we get
\[ \Rightarrow \] LHS \[ = \dfrac{{\cot A + \cos ecA - \left( {\cos e{c^2}A - {{\cot }^2}A} \right)}}{{\cot A + 1 - \cos ecA}}\]
Using the identity \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\], we get,
\[ \Rightarrow \] LHS \[ = \dfrac{{\cot A + \cos ecA - \left( {\cos ecA - \cot A} \right)\left( {\cos ecA + \cot A} \right)}}{{\cot A + 1 - \cos ecA}}\]
Factoring out common terms in the numerator, we get
\[ \Rightarrow \] LHS \[ = \dfrac{{\left( {\cos ecA + \cot A} \right)\left( {1 - \cos ecA + \cot A} \right)}}{{\cot A + 1 - \cos ecA}}\]
Cancelling out the same brackets from the numerator and the denominator,
\[ \Rightarrow \] LHS \[ = \cos ecA + \cot A = \] RHS

Therefore, \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A\]
Hence, proved

Note:
Here, we have considered the left side of the equation because it is easier to convert sine and cosine functions into other trigonometric functions. If we would have solved the equation from the right hand side then the solution would have been more complicated. There are six trigonometric functions but it has three basic trigonometric functions and they are sine, cosine and tangent functions. The other three functions are derived from these basic functions. The application of trigonometry is highly utilized in the field of engineering, architecture, etc. It helps to calculate the height and distance of objects and it is also used for navigation by ships and airplanes etc.