Question

prove that 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 you understood .