question_answer

Prove the identity: \[\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{1-2{{\cos }^{2}}A}\]

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.