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# Prove that $\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A$.

Last updated date: 16th Jun 2024
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Hint: For proving $\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A$, we divide the LHS part of the equation by $\sin A$ and convert this into $\dfrac{{\cot A - 1 + \cos ecA}}{{\cot A + 1 - \cos ecA}}$ and now we substitute identity in place of 1 i.e. $\cos e{c^2}A - {\cot ^2}A = 1$ in numerator after this we apply an identity in the numerator i.e. ${a^2} - {b^2} = (a + b)(a - b)$ after applying this identity we get terms as $\dfrac{{(\cos ecA + \cot A)(1 - \cos ecA + \cot A)}}{{\cot A + 1 - \cos ecA}}$ and we cancel out the like terms and we got our answer which is equal to the RHS.

By taking LHS $\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}}$ and we divide each term of numerator and denominator by $\sin A$.
We get,
$\dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}}}}$
After solving we get the equation as,
$\dfrac{{\cot A - 1 + \cos ecA}}{{\cot A + 1 - \cos ecA}}$
Now, in numerator in place of 1 we insert an identity which is,
$\cos e{c^2}A - {\cot ^2}A = 1$
We get the result as,
$\dfrac{{\cot A - (\cos e{c^2}A - {{\cot }^2}A) + \cos ecA}}{{\cot A + 1 - \cos ecA}}$
Now, we put an algebraic identity which is
${a^2} - {b^2} = (a + b)(a - b)$ where, $a = \cos ecA,b = \cot A$
We get,
$\dfrac{{\cos ecA + \cot A - [(\cos ecA + \cot A)(\cos ecA - \cot A)]}}{{\cot A + 1 - \cos ecA}}$
Now by taking $(\cos ecA + \cot A)$ we get the equation as,
$\dfrac{{(\cos ecA + \cot A)[1 - (\cos ecA - \cot A)]}}{{\cot A + 1 - \cos ecA}}$
By solving bracket we get
$\dfrac{{(\cos ecA + \cot A)[\cot A + 1 - \cos ecA]}}{{\cot A + 1 - \cos ecA}}$
By eliminating the like terms i.e. $\cot A + 1 - \cos ecA$
We get,
$\cos ecA + \cot A$
=RHS
Hence Proved

Note: Alternate Method to solve the above question
By taking LHS
$\dfrac{{\cos A - (\sin A - 1)}}{{\cos A + (\sin A - 1)}}$
And by dividing both numerator and denominator by $\cos A + \sin A - 1$
We get,
=$\dfrac{{\cos A - (\sin A - 1)}}{{\cos A + (\sin A - 1)}} \times \dfrac{{\cos A + (\sin A - 1)}}{{\cos A + (\sin A - 1)}}$
By applying identities (i) ${a^2} - {b^2} = (a + b)(a - b)$ in numerator, (ii) ${(a + b)^2} = {a^2} + {b^2} + 2ab$ in denominator
We get,
=$\dfrac{{{{\cos }^2}A - {{(\sin A - 1)}^2}}}{{{{[\cos A + (\sin A - 1)]}^2}}}$
=$\dfrac{{{{\cos }^2}A - {{(\sin A - 1)}^2}}}{{{{\cos }^2}A + {{(\sin A - 1)}^2} + 2\cos A \times (\sin A - 1)}}$
Now we apply identity ${(a - b)^2} = {a^2} + {b^2} - 2ab$ on $\sin A - 1$ in both numerator and denominator
We get,
=$\dfrac{{{{\cos }^2}A - ({{\sin }^2}A + 1 - 2\sin A)}}{{{{\cos }^2}A + {{\sin }^2}A + 1 - 2\sin A + 2\cos A\sin A - 2\cos A}}$
As, ${\cos ^2}A + {\sin ^2}A = 1$
=$\dfrac{{{{\cos }^2}A - {{\sin }^2}A - 1 + 2\sin A}}{{1 + 1 - 2\sin A + 2\cos A\sin A - 2\cos A}}$
=$\dfrac{{{{\cos }^2}A - (1 - {{\cos }^2}A) - 1 + 2\sin A}}{{2 - 2\sin A + 2\cos A\sin A - 2\cos A}}$
=$\dfrac{{{{\cos }^2}A - 1 + {{\cos }^2}A - 1 + 2\sin A}}{{2(1 - \sin A) + 2\cos A(\sin A - 1)}}$
=$\dfrac{{2{{\cos }^2}A - 2 + 2\sin A}}{{2(1 - \sin A) - 2\cos A(1 - \sin A)}}$
Taking (-2) common from numerator and $2(1 - \sin A)$ common from denominator
We get
=$\dfrac{{ - 2(1 - {{\cos }^2}A - \sin A)}}{{2(1 - \sin A)(1 - \cos A)}}$
=$\dfrac{{ - ({{\sin }^2}A - \sin A)}}{{(1 - \sin A)(1 - \cos A)}}$
Taking $\sin A$ common in numerator we get,
=$\dfrac{{ - \sin A(\sin A - 1)}}{{(1 - \sin A)(1 - \cos A)}}$
=$\dfrac{{\sin A(1 - \sin A)}}{{(1 - \sin A)(1 - \cos A)}}$
=$\dfrac{{\sin A}}{{(1 - \cos A)}}$
Now multiply and divide numerator and denominator by $(1 + \cos A)$
=$\dfrac{{\sin A}}{{(1 - \cos A)}} \times \dfrac{{(1 + \cos A)}}{{(1 + \cos A)}}$
We get
=$\dfrac{{\sin A \times (1 + \cos A)}}{{1 - {{\cos }^2}A}}$
=$\dfrac{{\sin A \times (1 + \cos A)}}{{{{\sin }^2}A}}$
We get
=$\dfrac{{1 + \cos A}}{{\sin A}}$
i.e. $\dfrac{1}{{\sin A}} + \dfrac{{\cos A}}{{\sin A}}$
=$\cos ecA + \cot A$
=RHS