Question

Evaluate $\sum \frac{\sin (A+B)\sin (A-B)}{\cos {}^{2}A\cos {}^{2}B}$

Answer 1

$\sum \frac{\sin (A+B)\sin (A-B)}{{\cos }^{2}A{\cos }^{2}B}$

$=\sum \frac{(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)}{{\cos }^{2}A{\cos }^{2}B}$

$=\sum \frac{({\sin }^{2}A{\cos }^{2}B-{\cos }^{2}A{\sin }^{2}B)}{{\cos }^{2}A{\cos }^{2}B}$

$=\sum \frac{\left({\sin }^{2}A{\cos }^{2}B\right)}{{\cos }^{2}A{\cos }^{2}B}-\sum \frac{\left({\cos }^{2}A{\sin }^{2}B\right)}{{\cos }^{2}A{\cos }^{2}B}$

$=\sum \frac{\left({\sin }^{2}A\right)}{{\cos }^{2}A}-\sum \frac{\left({\sin }^{2}B\right)}{{\cos }^{2}B}$

$=\sum {\tan }^{2}A-{\tan }^{2}B$

$={\tan }^{2}A-{\tan }^{2}B+{\tan }^{2}B-{\tan }^{2}C+{\tan }^{2}C-{\tan }^{2}A$

$=0$