Prove that frac{1+sec theta-tan theta}{1+sec | vedclass

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(A) $L$.$H$.$S$. $= \frac{1+\sec 	heta-	an 	heta}{1+\sec 	heta+	an 	heta}$$= \frac{1+\frac{1}{\cos 	heta}-\frac{\sin 	heta}{\cos 	heta}}{1+\f

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(A) $L$.$H$.$S$. $= \frac{1+\sec 	heta-	an 	heta}{1+\sec 	heta+	an 	heta}$$= \frac{1+\frac{1}{\cos 	heta}-\frac{\sin 	heta}{\cos 	heta}}{1+\frac{1}{\cos 	heta}+\frac{\sin 	heta}{\cos 	heta}}$ $\left[\because \sec 	heta=\frac{1}{\cos 	heta} 	ext{ and } 	an 	heta=\frac{\sin 	heta}{\cos 	heta}ight]$$= \frac{\cos 	heta+1-\sin 	heta}{\cos 	heta+1+\sin 	heta} = \frac{(\cos 	heta+1)-\sin 	heta}{(\cos 	heta+1)+\sin 	heta}$Using the identity $1 = \sec^2 	heta - 	an^2 	heta$,we can write:$= \frac{(\sec^2 	heta - 	an^2 	heta) + (\sec 	heta - 	an 	heta)}{1 + \sec 	heta + 	an 	heta}$$= \frac{(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) + (\sec 	heta - 	an 	heta)}{1 + \sec 	heta + 	an 	heta}$$= \frac{(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta + 1)}{1 + \sec 	heta + 	an 	heta}$$= \sec 	heta - 	an 	heta$$= \frac{1}{\cos 	heta} - \frac{\sin 	heta}{\cos 	heta} = \frac{1-\sin 	heta}{\cos 	heta} = 	ext{R.H.S.}$Hence proved.

Prove that $\frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}$

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