Question

If $\frac{cos(A-B)}{cos(A+B)}+\frac{cos(C+D)}{cos(C-D)}=0$, prove that $tanAtanBtanCtanD=-1$

Answer 1

We have,
$\frac{cos(A-B)}{cos(A+B)}+\frac{cos(C+D)}{cos(C-D)}=0,\Rightarrow \frac{cos(A-B)}{cos(A+B)}+\frac{cos(C+D)}{cos(C-D)}\dots \left(i\right)$
Now,
$\frac{cos(A-B)}{cos(A+B)}=-\frac{cos(C+D)}{cos(C-D)}\Rightarrow \frac{cos(A-B)}{cos(A+B)}+1=\frac{-cos(C+D)}{cos(C-D)}+1\Rightarrow \frac{cos(A-B)+cos(A+B)}{cos(A+B)}+\frac{cos(C+D)-cos(C-D)}{cos(C-D)}\Rightarrow \frac{cos(A+B)+cos(A-B)}{cos(A+B)}+\frac{cos(C+D)-cos(C-D)}{cos(C-D)}\left(ii\right)$
Again,
$\frac{cos(A-B)}{cos(A+B)}=\frac{-\left[cos\right(C+D)-cos(C-D\left)\right]}{cos(C-D)}\dots \left(ii\right)$
[By equation(i)]
$\Rightarrow \frac{cos(A-B)}{cos(A+B)}-1=\frac{-cos(C+D)}{cos(C-D)}-1\Rightarrow \frac{cos(A-B)-cos(A+B)}{cos(A+B)}=\frac{-\left[cos\right(C+D)+cos(C-D\left)\right]}{cos(C-D)}\Rightarrow \frac{cos(A+B)-cos(A-B)}{cos(A+B)}=\frac{cos(C+D)+cos(C-D)}{cos(C-D)}\left(iii\right)$
Dividing equation (ii) by equation (iii), we get
$\frac{cos(A+B)+cos(A-B)}{cos(A+B)-cos(A-B)}=\frac{-\left[cos\right(C+D)-cos(C-D\left)\right]}{cos(C+D)+cos(C-D)}\Rightarrow \frac{2cos\left\{\frac{A+B+A-B}{2}\right\}cos\left\{\frac{A+B-A+B}{2}\right\}}{2sin\left\{\frac{A+B+A-B}{2}\right\}sin\left\{\frac{A+B-A+B}{2}\right\}}$

$=\frac{-2sin\left[\left\{\frac{C+D+C-D}{2}\right\}sin\left\{\frac{C+D-C+D}{2}\right\}\right]}{2cos\left\{\frac{C+D+C-D}{2}\right\}cos\left\{\frac{C+D-C+D}{2}\right\}}\Rightarrow \frac{cosAcosB}{-sinAsinB}=\frac{sinCsinD}{-cosCcosD}\Rightarrow \frac{1}{tanAtanB}=tanCtanD\Rightarrow -1=tanAtanBtanCtanD\therefore tanAtanBtanCtanD=-1$