Question

If $\cos (A+B)\sin (C-D)=\cos (A-B)\sin (C+D),$ then the value of $\tan A\tan B\tan C+\tan D$ is

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is C $0$

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

By componendo and dividendo
$\Rightarrow \frac{\cos (A-B)+\cos (A+B)}{\cos (A-B)-\cos (A+B)}=\frac{\sin (C-D)+\sin (C+D)}{\sin (C-D)-\sin (C+D)}$
$\Rightarrow \frac{2\cos A\cos B}{2\sin A\sin B}=\frac{2\sin C\cos D}{-2\cos C\sin D}$
$\Rightarrow \cot A\cot B=-\tan C\cot D$
$\Rightarrow \tan A\tan B\tan C=-\tan D$
$\Rightarrow \tan A\tan B\tan C+\tan D=0$