回復 12# mathca 的帖子
第3題
\(\begin{align}
& \tan \alpha +\tan \beta =a \\
& \tan \alpha \tan \beta =b \\
& {{\cos }^{2}}\alpha -{{\sin }^{2}}\beta \\
& =\frac{1}{{{\sec }^{2}}\alpha }-\frac{1}{{{\csc }^{2}}\beta } \\
& =\frac{1}{1+{{\tan }^{2}}\alpha }-\frac{1}{1+{{\cot }^{2}}\beta } \\
& =\frac{1}{1+{{\tan }^{2}}\alpha }-\frac{{{\tan }^{2}}\beta }{1+{{\tan }^{2}}\beta } \\
& =\frac{1+{{\tan }^{2}}\beta -{{\tan }^{2}}\beta \left( 1+{{\tan }^{2}}\alpha \right)}{\left( 1+{{\tan }^{2}}\alpha \right)\left( 1+{{\tan }^{2}}\beta \right)} \\
& =\frac{1-{{\tan }^{2}}\alpha {{\tan }^{2}}\beta }{\left( 1+{{\tan }^{2}}\alpha \right)\left( 1+{{\tan }^{2}}\beta \right)} \\
& =\frac{1-{{\left( \tan \alpha \tan \beta \right)}^{2}}}{1+{{\left( \tan \alpha +\tan \beta \right)}^{2}}-2\tan \alpha \tan \beta +{{\left( \tan \alpha \tan \beta \right)}^{2}}} \\
& =\frac{1-{{b}^{2}}}{1+{{a}^{2}}-2b+{{b}^{2}}} \\
& =\frac{1-{{b}^{2}}}{{{a}^{2}}+{{\left( b-1 \right)}^{2}}} \\
\end{align}\)