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

We can use the trigonometric identity (sec^{2}θ = 1 + tan^{2}θ)

## 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) + (sec^{2}θ - tan^{2}θ)} (Since, 1 =sec^{2}θ - tan^{2}θ)

= {( 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

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