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Prove that, ...

Question

Prove that,
$\frac{sin A}{1 - cos A} = \frac{1 + cos A - sin A}{sin A - 1 + cos A}$

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Solution

$\frac{sin A}{1 - cos A} = \frac{1 + cos A - sin A}{sin A - 1 + cos A}$
$\frac{1 + cos A - sin A}{sin A - 1 + cos A}$
$\Rightarrow$ divide by $sin A$
$\Rightarrow \frac{csc A + cot A - 1}{1 - csc A + cot A}$
using $1 + {cot}^{2} A = {csc}^{2} A$
$\Rightarrow \frac{csc A + cot A - 1}{({csc}^{2} A - {cot}^{2} A) - csc A + cot A}$
$\Rightarrow \frac{[csc A + cot A - 1]}{- (cot A - csc A) [csc A + cot A - 1]}$
$\Rightarrow \frac{- 1}{cot A - csc A}$
$\Rightarrow \frac{- sin A}{cos A - 1}$
$\Rightarrow \frac{sin A}{1 - cos A}$

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\frac{sin A + 1 - cos A}{sin A - 1 + cos A} = \frac{1 + sin A}{cos A}

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\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{s i n^{2} A - c o s^{2} A} = \frac{2}{2 s i n^{2} A - 1} = \frac{2}{1 - 2 c o s^{2} A}

\frac{cos A}{1 + sin A} + \frac{1 + sin A}{cos A} = 2 sec A

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