Question

sin theta-cos theta+1/sin theta+cos theta-1=1/sec theta-tan theta prove

Answer 1

LHS = (sinθ - cosθ + 1)/(sinθ + cosθ - 1)
dividing by cosθ both Numerator and denominator
= (sinθ/cosθ - cosθ/cosθ + 1/cosθ)/(sinθ/cosθ + cosθ/cosθ - 1/cosθ)
= (tanθ + secθ - 1)/(tanθ - secθ + 1)

Multiply (tanθ - secθ) with both Numerator and denominator
= (tanθ + secθ - 1)(tanθ - secθ)/(tanθ - secθ + 1)(tanθ - secθ)
= {(tan²θ - sec²θ) - (tanθ - secθ)}/(tanθ - secθ + 1)(tanθ - secθ)
= (-1 - tanθ + secθ)/(tanθ - secθ + 1)(tanθ - secθ) [ ∵ sec²x - tan²x = 1 ]
= -1/(tanθ - secθ) = 1/(secθ - tanθ) = RHS

Answer 2

Answer:

Step-by-step explanation:

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