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

\[\begin{aligned} \mathrm{LHS} & =\frac{(\tan \theta+\sec \theta)-\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{\tan \theta-\sec \theta+1} \\ & =\frac{(\tan \theta+\sec \theta)(1-\sec \theta+\tan \theta)}{\tan \theta-\sec \theta+1} \\ & =\tan \theta+\sec \theta \\ = & \frac{1+\sin \theta}{\cos \theta}=\mathrm{RHS} \end{aligned} \]