Question

In $\Delta ABC,\sum \frac{sin(A+B).sin(A-B)}{{cos}^{2}A{cos}^{2}B}=$

• A. 0
• B. 1
• C. 2
• D. 1/2

Answer 1

Answer: (A)

0

Consider $\sin (A+B)\sin (A-B)=(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)$

$={\sin }^{2}A{\cos }^{2}B-{\cos }^{2}A{\sin }^{2}B$

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

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

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

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

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

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

$=0$