Question

pls ! any body prove this.
(sin theta-cos
theta+1)/(sin theta+cos theta-1)=1/(sec theta-tan theta)

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



$hope \: it \: will \: help \: you$