Question

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

Answer 1

Hint: We take LHS and simplify it to get RHS. We divide the numerator and the denominator by Sine function. We know that cotangent is the ratio of cosine to sine function and using the relation between cosine and cotangent we can solve this.

Complete step by step solution:
Given
\[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \csc A + \cot A\]
Now take
\[LHS = \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}}\] and \[RHS = \csc A + \cot A\].
Now take LHS,
\[LHS = \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}}\]
Now divide the numerator and denominator by \[\sin A\]
\[LHS = \dfrac{{\left( {\dfrac{{\cos A - \sin A + 1}}{{\sin A}}} \right)}}{{\left( {\dfrac{{\cos A + \sin A - 1}}{{\sin A}}} \right)}}\]
\[LHS = \dfrac{{\left( {\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}} \right)}}{{\left( {\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}}} \right)}}\]
\[LHS = \dfrac{{\left( {\dfrac{{\cos A}}{{\sin A}} - 1 + \dfrac{1}{{\sin A}}} \right)}}{{\left( {\dfrac{{\cos A}}{{\sin A}} + 1 - \dfrac{1}{{\sin A}}} \right)}}\]
We know \[\dfrac{{\cos A}}{{\sin A}} = \cot A\] and \[\dfrac{1}{{\sin A}} = \csc A\]then we have,
\[LHS = \dfrac{{\cot A - 1 + \csc A}}{{\cot A + 1 - \csc A}}\]
We know \[{\csc ^2}A - {\cot ^2}A = 1\], applying this in the numerator we have,
\[LHS = \dfrac{{\cot A + \csc A - \left( {{{\csc }^2}A - {{\cot }^2}A} \right)}}{{\cot A + 1 - \csc A}}\]
Now applying \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] then we have,
\[LHS = \dfrac{{\cot A + \csc A - \left( {\left( {\csc A - \cot A} \right)\left( {\csc A + \cot A} \right)} \right)}}{{\cot A + 1 - \csc A}}\]
Now raking \[\csc A + \cot A\] common we have,
\[LHS = \dfrac{{\left( {\csc A + \cot A} \right)\left( {1 - \left( {\csc A - \cot A} \right)} \right)}}{{\cot A + 1 - \csc A}}\]
\[LHS = \dfrac{{\left( {\csc A + \cot A} \right)\left( {1 - \csc A + \cot A} \right)}}{{\cot A + 1 - \csc A}}\]
Now cancelling the terms we have,
\[LHS = \csc A + \cot A\]
\[ \Rightarrow LHS = RHS\].
Thus \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \csc A + \cot A\]. Hence proved.

Note: Remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.