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Question
Prove that:
\[\frac{\sin A + 1 - \cos A}{\cos A - 1 + \sin A} = \frac{1 + \sin A}{\cos A}\]
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Solution
Solving L.H.S. of the equation:
\[\Rightarrow \frac{1 + \sin A - \cos A}{\sin A + \cos A - 1}\]
Multiplying numerator and denominator by (1 + sin A + cos A) :
\[\Rightarrow \frac{(1 + \sin A - \cos A)(1 + \sin A + \cos A)}{(\sin A + \cos A - 1)(\sin A + \cos A + 1)}\]
\[\Rightarrow \frac{(1 + \sin A)^2 - \cos^2 A}{(\sin A + \cos A)^2 - 1}\]
\[\Rightarrow \frac{1 + 2 \sin A + \sin^2 A - \cos^2 A}{\sin^2 A + \cos^2 A + 2 \sin A \cos A - 1}\]
By formula,
\[\cos^2 A = 1 - \sin^2 A\] and \[\sin^2 A + \cos^2 A = 1.\]
\[\Rightarrow \frac{1 + 2 \sin A + \sin^2 A - (1 - \sin^2 A)}{1 + 2 \sin A \cos A - 1}\]
\[\Rightarrow \frac{2 \sin A + 2 \sin^2 A}{2 \sin A \cos A}\]
\[\Rightarrow \frac{2 \sin A (1 + \sin A)}{2 \sin A \cos A}\]
\[\Rightarrow \frac{1 + \sin A}{\cos A}.\]
Since, L.H.S. = R.H.S.
