Question

Prove:-
$\sf \dfrac{sinA \: - cosA + 1 }{sinA \: + cosA - 1} = \dfrac{1}{secA - tanA}$
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Answer 1

Answer:

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Answer 2

$LHS = \frac{sinA - cosA + 1}{sinA + cosA - 1}$

⇒ Divide cosA in each term

⇒ $\frac{tanA -1 + secA}{tanA +1 - secA}$

⇒ $\frac{(tanA + secA )- 1}{(tanA - secA) + 1}$

⇒ $\frac{[ (tanA + secA ) -1] ( tanA - secA )}{ [(tanA - secA ) +1] ( tanA - secA)}$

⇒ $\frac{( tan^{2}A - sec^{2}A) - ( tanA - secA) }{( tanA - secA + 1 )( tanA - secA)}$

⇒ $\frac{-1-tanA+secA}{( tanA - secA + 1) (tanA - sec A)}$

⇒ $\frac{-1}{tanA - secA}$

⇒ $\frac{1}{secA - tanA}$

So LHS = RHS

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