${{\sin }^{4}}\alpha +{{\cos }^{4}}\alpha ={{({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha )}^{2}}-2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha =1-2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \,\,\,\,\,\,\,(1)$

$\tan \alpha +\cot \alpha =\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }=\frac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }{\sin \alpha \cos \alpha }=\frac{1}{\sin \alpha \cos \alpha }\,\,\,\,\,\,\,\,\,(2)$

با جایگذاری روابط (۱) و (۲) در $A$ داریم:

$A=1-2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha +\frac{2\sin \alpha \cos \alpha }{\frac{1}{\sin \alpha \cos \alpha }}=1-2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha +2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha =1$

راه حل دیگر:

\[\tan \,\alpha +\cot \alpha =\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }=\frac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha }{\sin \alpha \cos \alpha }=\frac{1}{\sin \alpha \cos \alpha }\]

با جایگذاری در $A$ داریم:

${{\sin }^{4}}\alpha +{{\cos }^{4}}\alpha +\frac{2\sin \alpha \cos \alpha }{\frac{1}{\sin \alpha \cos \alpha }}={{\sin }^{4}}\alpha +{{\cos }^{4}}\alpha +2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha ={{({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha )}^{2}}={{1}^{2}}=1$