Prove that:
$$\frac{\sin heta - \cos heta + 1}{\cos heta + \sin heta - 1} = \frac{1}{\sec heta - an heta}$$

Question

Prove that:
$$\frac{\sin 	heta - \cos 	heta + 1}{\cos 	heta + \sin 	heta - 1} = \frac{1}{\sec 	heta - 	an 	heta}$$

Accepted Answer

Step 1: L.H.S $= \frac{\sin 	heta - \cos 	heta + 1}{\cos 	heta + \sin 	heta - 1}$. Dividing numerator and denominator by $\cos 	heta$: $= \frac{	an 	heta - 1 + \sec 	heta}{1 + 	an 	heta - \sec 	heta}$
Step 2: $= \frac{	an 	heta + \sec 	heta - 1}{1 + 	an 	heta - \sec 	heta}$. Using $\sec^2 	heta - 	an^2 	heta = 1$: $= \frac{(\sec 	heta - 	an^2 	heta) + 	an 	heta - \sec 	heta}{\sec 	heta - 	an 	heta(\sec 	heta + 	an 	heta - 1)}$
Step 3: $= \frac{	an 	heta + \sec 	heta - 1}{(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta - 1)} = \frac{1}{\sec 	heta - 	an 	heta} = $ R.H.S

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