Question

Prove that $\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A$, using the identity $\cos e{c^2}A = 1 + {\cot ^2}A$.

Answer 1

Hint:
Here we will take the left hand side of the given expression and we will divide the numerator and denominator by $\sin A$. Then we will modify the given identity to substitute in the obtain equation. Then we also use the algebraic identity to simplify it. On further simplifying the terms, we will get $\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A$.

Complete step by step solution:
We take the left hand side of the equation and we will prove that the left hand side of the equation is equal to the right hand side of the equation.
On dividing the numerator and denominator by $\sin A$, we get
$\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}}}}$
We know, $\dfrac{{\cos A}}{{\sin A}} = \cot A$ and $\dfrac{1}{{\sin A}} = \cos ecA$.
Substituting these values in the equation, we get
$ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\cot A - 1 + \cos ecA}}{{\cot A + 1 - \cos ecA}}$
We know from trigonometric identities; $\cos e{c^2}A - {\cot ^2}A = 1$.
So we will substitute $\cos e{c^2}A - {\cot ^2}A$ in place of 1 in the numerator.
$ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\cot A + \cos ecA - \left( {\cos e{c^2}A - {{\cot }^2}A} \right)}}{{\cot A + 1 - \cos ecA}}$
Using the algebraic identity; ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ in the above equation, we get
$ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\cot A + \cos ecA - \left( {\cos ecA + \cot A} \right)\left( {\cos ecA - \cot A} \right)}}{{\cot A + 1 - \cos ecA}}$
Simplifying the terms in numerator, we get
$ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\left( {\cot A + \cos ecA} \right)\left( {1 - \cos ecA + \cot A} \right)}}{{\cot A + 1 - \cos ecA}}$
On further simplifying the terms of numerator and denominator, we get
$ \Rightarrow \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cot A + \cos ecA$
Hence, we have proved the equation $\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A$.

Note:
To prove any trigonometric identity, we need to choose the side of the equation which can be easily converted and modified into terms present on the other side. We can also take the right side of the equation and simplify it to prove the given equation. But it will be much more difficult to solve, so we did not solve the right side of the equation. We need to remember the basic trigonometric formulas and identities to solve this question.