Show that (sinθ  - cosθ + 1) / ( sinθ + cosθ - 1) = 1 /(secθ - tanθ )

We can use the trigonometric identity  (sec²θ = 1 + tan²θ)

Answer : (sinθ  - cosθ + 1) / ( sinθ + cosθ - 1) = 1 /(secθ - tanθ )

Let us deduce the expression from LHS towards RHS

Explanation:

Since we will apply the identity involving secθ and tanθ, first convert the LHS in terms of secθ and tanθ by dividing both numerator and denominator by cosθ.

thus,

LHS = ( sinθ - cosθ +1) / (sinθ + cosθ -1)

         = (sinθ/cosθ - cosθ/cosθ +1/cosθ) / (sinθ/cosθ + cosθ/cosθ -1/cosθ)

        = (tanθ - 1 + secθ) / ( tanθ + 1 - secθ)                              (Since, sinθ/cosθ = tanθ and 1/cosθ = secθ)

        = {( tanθ + secθ) -1} / {(tanθ - secθ) +1} 

       =  {( tanθ + secθ) -1} / {(tanθ - secθ) + (sec²θ - tan²θ)}         (Since, 1 =sec²θ - tan²θ)

      =  {( tanθ + secθ) -1} / {(tanθ - secθ) + (secθ + tanθ)(secθ - tanθ)}

       =  {( tanθ + secθ) -1} / {-(secθ - tanθ)+ (secθ + tanθ)(secθ - tanθ)}

        = {( tanθ + secθ) -1} / {(secθ - tanθ)[-1+ (secθ + tanθ)]}

        = {( tanθ + secθ) -1} / {(secθ - tanθ)( tanθ + secθ -1)}

        = 1 / (secθ - tanθ)

       Hence, proved LHS = RHS