Question

${ \rm{ \huge \underline{Prove \: It }}}$


${ \rm{\dfrac{sin \: A - cos \: A + 1}{sin \: A + cos \: A - 1} = \dfrac{cos \: A}{1 - sin \:A} }}$

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Answer 1

Solution:-

$\rm \implies \dfrac{ \sin \: A - \cos A + 1}{\sin \: A + \cos A - 1} = \dfrac{ \cos A }{1 - \sin A }$

Using cross multiplication method

$\rm \: \implies(1 - \sin A)( \sin A - \cos A + 1) = \cos A( \sin A + \cos A - 1)$

$\rm \implies \: \sin A - \cos A + 1 - \sin {}^{2} A + \cos A \sin A - \sin A = \\ \rm\cos A \sin A + \cos {}^{2} A - \cos A$

$\rm \implies \: \sin A - \cos A + 1 - \sin {}^{2} A + \cancel{ \cos A \sin A} - \sin A = \\ \rm \cancel{\cos A \sin A} + \cos {}^{2} A - \cos A$

$\rm \implies \: \sin A - \cos A + 1 - \sin {}^{2} A - \sin A = \rm \cos {}^{2} A - \cos A$

$\rm \implies \: \cancel{ \sin A } - \cancel{\cos A} + 1 - \sin {}^{2} A - \cancel{\sin A }= \rm \cos {}^{2} A - \cancel{\cos A}$

$\rm \implies \: 1 - \sin {}^{2} A = \cos {}^{2} A$

$\rm \implies \: 1 = \cos {}^{2} A + \sin {}^{2} A$

$\implies1 = 1$

LHS = RHS

Hence proved