Question

prove that $\dfrac{sinθ-cosθ+1}{sinθ+cosθ-1}$ = $\dfrac{1}{secθ - tanθ}$ ​

Answer 1

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Answer 2

$\huge\underline\bold\blue{solution:-}$

⟹ $\dfrac{sinθ - cosθ + 1}{sinθ + cosθ - 1}$

⟹ $\dfrac{tanθ + 1 - secθ}{tanθ + 1 - secθ}$

⟹ $\dfrac{(tanθ + secθ) - 1}{(tanθ - secθ) + 1}$

⟹ $\dfrac{{(tanθ + secθ) - 1} (tanθ - secθ)}{{(tanθ - secθ) + 1} ( tanθ - secθ)}$

⟹ $\dfrac{(tan²θ - sec²θ) - (tanθ - secθ}{{(tanθ - secθ) + 1} ( tanθ - secθ)}$

⟹ $\dfrac{-1 - tanθ + secθ}{{(tanθ - secθ) + 1} ( tanθ - secθ)}$

⟹ $\dfrac{-1}{tanθ - secθ}$

⟹ $\dfrac{1}{secθ - tanθ}$

Hence

$\underline\bold\purple{L.H.S}$ =$\underline\bold\purple{R.H.S}$

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