Question

Solve the following: $\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A$.

Answer 1

In the given question , we will solve the expression on the LHS and prove it equal to the RHS expression . The expression in LHS contains $\cos$ and $\sin$ , so we will convert them to $\cos ec$ and $\cot$ by dividing the numerator and denominator by $\sin A$ . Also we will be using a basic trigonometric identity i.e. $\cos e{c^2}A = 1 + {\cot ^2}A$.

Complete step by step answer:
Given : $\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}}$
Now on dividing the numerator and denominator by $\sin A$ we get ,
$LHS = \dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}}}}$
On solving the above expression as $\dfrac{1}{{\sin A}} = \cos ecA$ we get ,
$LHS = \dfrac{{\cot A - 1 + \cos ecA}}{{\cot A + 1 - \cos ecA}}$
Now in numerator we will use the trigonometric identity $\cos e{c^2}A = 1 + {\cot ^2}A$ , we get
$LHS = \dfrac{{\cot A + \cos ecA - \left( {\cos e{c^2}A - {{\cot }^2}A} \right)}}{{\cot A + 1 - \cos ecA}}$
Now using the identity of ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ , we get
$LHS= \dfrac{{\cot A + \cos ecA - \left[ {\left( {\cos ecA - \cot A} \right)\left( {\cos ecA + \cot A} \right)} \right]}}{{\cot A + 1 - \cos ecA}}$

Now taking $\left( {\cos ecA + \cot A} \right)$ common in numerator , we get
$LHS = \dfrac{{\left( {\cot A + \cos ecA} \right)\left[ {1 - \left( {\cos ecA + \cot A} \right)} \right]}}{{\cot A + 1 - \cos ecA}}$
On rearranging the terms in numerator we get ,
$LHS= \dfrac{{\left( {\cot A + \cos ecA} \right)\left[ {\cot A + 1 - \cos ecA} \right]}}{{\cot A + 1 - \cos ecA}}$
Now cancelling out the term $\cot A + 1 - \cos ecA$ from numerator and denominator , we get
$LHS = \left( {\cot A + \cos ecA} \right)$
Hence Proved. Therefore , the $LHS = RHS$ .

Note: Alternate Method :
$LHS = \dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}}$
Now multiplying the numerator and denominator by $\sin A$ we get ,
$LHS = \dfrac{{\sin A\left( {\cos A - \sin A + 1} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}$
On solving we get ,
$LHS = \dfrac{{\sin A\cos A - {{\sin }^2}A + \sin A}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}$
Now using the basic trigonometric identity ${\sin ^2}A + {\cos ^2}A = 1$ in numerator we get ,
$LHS= \dfrac{{\sin A\cos A + \sin A - \left( {1 - {{\cos }^2}A} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}$
Now using the identity of ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ , we get
$LHS = \dfrac{{\sin A\cos A + \sin A - \left[ {\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)} \right]}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}$
Now taking $\sin A$ common in numerator we get ,
$LHS= \dfrac{{\sin A\left( {\cos A + 1} \right) - \left[ {\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)} \right]}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}$

Now taking $\left( {\cos A + 1} \right)$ common from the numerator we get ,
$LHS= \dfrac{{\left( {\cos A + 1} \right)\left[ {\sin A - \left( {1 - \cos A} \right)} \right]}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}$
On rearranging the terms in numerator we get ,
$LHS = \dfrac{{\left( {\cos A + 1} \right)\left( {\cos A + \sin A - 1} \right)}}{{\sin A\left( {\cos A + \sin A - 1} \right)}}$
Now cancelling out the term $\left( {\cos A + \sin A - 1} \right)$ from numerator and denominator we get ,
$LHS = \dfrac{{\left( {\cos A + 1} \right)}}{{\sin A}}$
On simplifying we get ,
$LHS= \dfrac{{\cos A}}{{\sin A}} + \dfrac{1}{{\sin A}}$
On solving as $\dfrac{1}{{\sin A}} = \cos ecA$ we get ,
$LHS = \cot A + \cos ecA$
Hence Proved.