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

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