Question

If $\frac{\cos (A+B)}{\cos (A-B)}+\frac{\cos (C-D)}{\cos (C+D)}=0$, then $\cot A\cot B\cot C$ is equal to

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

Answer 1

The correct option is C $-\tan D$

Given, $\frac{\cos (A+B)}{\cos (A-B)}+\frac{\cos (C-D)}{\cos (C+D)}=0$

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

$\Rightarrow \frac{1-\tan A\tan B}{1+\tan A\tan B}+\frac{1+\tan C\tan D}{1-\tan C\tan D}=0$

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

$\Rightarrow \tan A\tan B\tan C\tan D=-1$

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