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Trigonometric Identity- 1

Prove that ...

Question

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

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Solution

$\frac{S i n A - (c o s A - 1)}{S i n + (c o s - 1)} \times \frac{S i n A - (c o s A - 1)}{S i n A - (c o s A - 1)}$

$\Rightarrow \frac{S i n^{2} A + (c o s A - 1)^{2} - 2 s i n A (c o s A - 1)}{s i n^{2} A - (c o s A - 1)^{2}}$

$\Rightarrow \frac{1 + 1 - 2 c o s A - 2 s i n A c o s A + 2 s i n A}{(2 - 2 c o s A) c o s A}$

$\Rightarrow \frac{2 (1 - c o s A) + 2 s i n A (1 - c o s A)}{2 (1 - c o s A) c o s A}$

$\Rightarrow \frac{1 + s i n A}{c o s A} \times \frac{1 - s i n A}{1 - s i n A} = \frac{c o s^{2} A}{c o s A (1 - s i n A)}$

$\Rightarrow \frac{c o s A}{1 - s i n A}$

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