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

Question

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

Accepted Answer

Step 1: LHS $= \frac{(\sec^2 	heta - 	an^2 	heta) + (\sec 	heta - 	an 	heta)}{1 + \sec 	heta + 	an 	heta} = \frac{(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta + 1)}{1 + \sec 	heta + 	an 	heta}$
Step 2: $= \sec 	heta - 	an 	heta$
Step 3: $= \frac{1}{\cos 	heta} - \frac{\sin 	heta}{\cos 	heta} = \frac{1 - \sin 	heta}{\cos 	heta} = $ RHS

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