question_answer

If \[\frac{\cos \,B}{\sin \,A}=n\] and \[\frac{\cos \,B}{\cos \,A}=m\] then show that \[({{m}^{2}}+{{n}^{2}})co{{s}^{2}}A={{n}^{2}}\].

Answer:

Given,               \[n=\frac{\cos \,B}{\sin \,A};\,\,m=\frac{\cos \,B}{\cos \,A}\]

So,                   \[{{n}^{2}}=\frac{{{\cos }^{2}}\,B}{{{\sin }^{2}}\,A};\,\,{{m}^{2}}=\frac{{{\cos }^{2}}\,B}{{{\cos }^{2}}\,A}\]

\[L.H.S.=({{m}^{2}}+{{n}^{2}})co{{s}^{2}}A=\left( \frac{{{\cos }^{2}}B}{{{\cos }^{2}}A}+\frac{{{\cos }^{2}}B}{{{\sin }^{2}}A} \right){{\cos }^{2}}A\]

\[=\frac{(si{{n}^{2}}A\,\,co{{s}^{2}}B+co{{s}^{2}}A\,\,co{{s}^{2}}B)}{{{\cos }^{2}}A\,\,{{\sin }^{2}}A}\times {{\cos }^{2}}A\]

\[=\frac{{{\cos }^{2}}B(si{{n}^{2}}A+co{{s}^{2}}A)}{{{\sin }^{2}}A}\]

\[=\frac{{{\cos }^{2}}B}{{{\sin }^{2}}A}\]

\[={{n}^{2}}=R.H.S.\]                                  Hence Proved.