Question

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

Answer 1

Hint: 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.