Answer:
\[L.H.S.=\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}\] \[=\frac{{{(\sin \,A+\cos \,A)}^{2}}+{{(\sin \,A-\cos \,A)}^{2}}}{(\sin A-\cos A)(\sin A+\cos A)}\] \[=_{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\sin }^{2}}A\,\,-\,\,{{\cos }^{2}}A\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}^{\underline{{{\sin }^{2}}A\,+\,\,{{\cos }^{2}}A\,\,+\,\,2\sin A\cos A+\,\,{{\sin }^{2}}A\,\,+\,\,{{\cos }^{2}}A\,\,-\,\,2\sin A\cos A}}\] \[=\frac{1+1}{1-{{\cos }^{2}}A-{{\cos }^{2}}A}\] \[\begin{align} & [\because \,{{\sin }^{2}}A+{{\cos }^{2}}A=1, \\ & \,\,\,\,\,{{\sin }^{2}}A=1-{{\cos }^{2}}A] \\ \end{align}\] \[=\frac{2}{1-2\,{{\cos }^{2}}A}=R.H.S\] Hence Proved.
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