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.
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