Question

$\frac{\cos (A-B)}{\cos (A+B)}+\frac{\cos (C+D)}{\cos (C-D)}=0\Rightarrow \tan A\tan B\tan C\tan D$ is equal to

• A. $0$
• B. 1
• C. $-1$
• D. 2

Answer 1

The correct option is C $-1$

$\frac{cos(A-B)}{cos(A+B)}+\frac{cos(C+D)}{cos(C-D)}=\frac{1+tanAtanB}{1-tanAtanB}+\frac{1-tanCtanD}{1+tanCtanD}$
$=\frac{1+tanAtanB+tanCtanD+tanAtanBtanCtanD+1-tanAtanB-tanCtanD+tanAtanBtanCtanD}{(1-tanAtanB)(1+tanCtanD)}$
$=\frac{2(1+tanAtanBtanCtanD)}{(1-tanatanB)(1+tanCtanD)}$
Given $\frac{cos(A-B)}{cos(A+B)}+\frac{cos(C+D)}{cos(C-D)}=0$
$tanAtanBtanCtanD=-1$