If $\displaystyle \sec \theta+\tan \theta=\mathrm{m}$, show that $\displaystyle \frac{\mathrm{m}^{2}-1}{\mathrm{~m}^{2}+1}=\sin \theta$.

L.H.S. \(\displaystyle =\frac{m^{2}-1}{m^{2}+1}\) \[\begin{aligned} & =\frac{(\sec \theta+\tan \theta)^{2}-\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{(\sec \theta+\tan \theta)^{2}+\left(\sec ^{2} \theta-\tan ^{2} \theta\right)} \\ & =\frac{(\sec \theta+\tan \theta)(\sec \theta+\tan \theta-\sec \theta+\tan \theta)}{(\sec \theta+\tan \theta)(\sec \theta+\tan \theta+\sec \theta-\tan \theta)} \\ & =\frac{\tan \theta}{\sec \theta} \\ & =\sin \theta=\text { R.H.S. } \end{aligned} \]