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red_dog · #2 July 3rd, 2009, 11:32 PM

$\frac{1}{\sin A}+\frac{1}{\cos A}=\frac{1}{\sin B}+\frac{1}{\cos B}\Rightarrow$

$\Rightarrow\frac{1}{\sin A}-\frac{1}{\sin B}=\frac{1}{\cos B}-\frac{1}{\cos A}\Rightarrow$

$\frac{\sin B-\sin A}{\sin A\sin B}=\frac{\cos A-\cos B}{\cos A\cos B}\Rightarrow\frac{\sin A\sin B}{\cos A\cos B}=\frac{\sin B-\sin A}{\cos A-\cos B}\Rightarrow$

$\tan A\tan B=\frac{2\sin\frac{B-A}{2}\cos\frac{A+B}{2}}{2\sin\frac{B-A}{2}\sin\frac{A+B}{2}}=\cot\frac{A+B}{2}$

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