已知 $\dfrac{\sin A}{\cos B}+\dfrac{\sin B}{\cos A}=2$，求证：$A+B=\dfrac{\pi}2$．
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【答案】
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【解析】
根据已知\[\begin{split}\dfrac{{\sin A}}{{\cos B}} + \dfrac{{\sin B}}{{\cos A}} = 2&\Leftrightarrow \sin A\cos A + \sin B\cos B = 2\cos A\cos B\\&\Leftrightarrow \dfrac{1}{2}\sin 2A + \dfrac{1}{2}\sin 2B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\\&\Leftrightarrow \sin \left( {A + B} \right)\cos \left( {A - B} \right) = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\\&\Leftrightarrow \left[ {1 - \sin \left( {A + B} \right)} \right]\cos \left( {A - B} \right) = - \cos \left( {A + B} \right)\\&\Leftrightarrow {\left[ {1 - \sin \left( {A + B} \right)} \right]^2}{\cos ^2}\left( {A - B} \right) = 1 - {\sin ^2}\left( {A + B} \right),\end{split}\]而$${\left[ {1 - \sin \left( {A + B} \right)} \right]^2} \leqslant 1 - {\sin ^2}\left( {A + B} \right),$$等号当且仅当 $\sin \left( {A + B} \right) = 1$ 时取得．于是 $A + B = \dfrac{{\rm{\pi }}}{2}$．
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