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

Solution $$\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$$ $$=\frac{(\sin\theta-\cos\theta+1)(\sin\theta+\cos\theta+1)}{(\sin\theta+\cos\theta-1)(\sin\theta+\cos\theta+1)}$$ $$=\frac{(\sin\theta+1-\cos\theta)(\sin\theta+1+\cos\theta)}{(\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1)}$$ $$=\frac{(\sin\theta+1)^2-\cos^2\theta}{(\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1)}$$ Since $$\sin^2\theta+\cos^2\theta=1$$, we get $$\frac{(1-\cos^2\theta+2\sin\theta)}{(2\sin\theta\cos\theta)}$$ $$=\frac{(2-2\cos^2\theta+2\sin\theta)}{(2\sin\theta\cos\theta)}$$ $$=\frac{(1-\cos^2\theta+\sin\theta)}{(sin\theta\cos\theta)}$$ $$=\frac{(\sin\theta+\sin^2\theta)}{(\sin\theta\cos\theta)}$$ $$=\frac{(\sin\theta+1)}{\cos\theta}$$ $$=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}$$ $$=\sec\theta+\tan\theta$$ $$=(\sec\theta+\tan\theta)\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}$$ $$=\frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}$$ We know that$$\sec^2\theta-\tan^2\theta=1$$ and so we get $$=\frac{1}{\sec\theta-\tan\theta}$$ Hence, proved.