Question

If $\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$, prove that $\tan A\tan B\tan C+\tan D=0$

Answer 1

$\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$

Using compounds dividend

$\frac{\cos (A+B)+\cos (A-B)}{\cos (A+B)-\cos (A-B)}=\frac{\sin (C+D)+\sin (C-D)}{\sin (C+D)-\sin (C-D)}$

$\frac{2\cos A\cdot \cos B}{-2\sin A\cdot \sin B}=\frac{2\sin C\cdot \cos D}{2\cos C\cdot \sin D}$

$-\cot A\cdot \cot B=\tan C\cdot \cot D$

$-\frac{1}{\tan A\cdot \tan B}=\frac{\tan C}{\tan D}$

$\therefore -\tan D=\tan A\cdot \tan B\cdot \tan C$

$\therefore \tan A\cdot \tan B\cdot \tan C+\tan D=0$.