Question

If $\frac{\cos (A-B)}{\cos (A+B)}+\frac{\cos (C+D)}{\cos (C-D)}=0$, then $\tan AtanB\tan C=$

• A. $\tan D$
• B. $\cot D$
• C. $-\tan D$
• D. $-\cot D$

Answer 1

The correct option is D $-\cot D$

$\frac{\cos (A-B)}{\cos (A+B)}+\frac{\cos (C+D)}{\cos (C-D)}=\frac{1+\tan A\tan B}{1-\tan A\tan B}+\frac{1-\tan C\tan D}{1+\tan C\tan D}$

$\left[\cos \right(A-B)=\cos A\cos B+\sin A\sin B,\cos (A+B)=\cos A\cos B-\sin A\sin B]$

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

$=\frac{2+2\tan A\tan B\tan C\tan D}{(1+\tan C\tan D)(1-\tan A\tan B)}=0$

$\tan A\tan B\tan C\tan D+1=0$

$\Rightarrow \tan A\tan B\tan C=-cotD$