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sec + tan -1/ tan -sec +1=cos/1- sin

Taking LHS, \frac{secx + tanx - 1}{tanx - secx + 1} = \frac{\frac{1}{cosx} + \frac{sinx}{cosx} - 1}{\frac{sinx}{cosx} - \frac{1}{cosx} + 1} = \frac{1 + sinx - cosx}{sinx - 1 + cosx} = \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2} - 1 + 2 \sin^{2} (\frac{x}{2})}{2 \sin \frac{x}{2} \cos \frac{x}{2} - 1 + 1 - 2 \sin^{2} (\frac{x}{2})} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \times \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} = \frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} - 2 \cos \frac{x}{2} \sin \frac{x}{2}} = \frac{cosx}{1 - sinx}

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