Question

If $cos(A+B)sin(C-D)=cos(A-B)sin(C+D),provethattanAtanBtanC+tanD=0$

Answer 1

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

$\Rightarrow \frac{2cos\left\{\frac{A+B+A-B}{2}\right\}cos\left\{\frac{A+B-A+B}{2}\right\}}{-2sin\left\{\frac{A+B+A-B}{2}\right\}sin\left\{\frac{A+B-A+B}{2}\right\}}=\frac{2sin\left\{\frac{C+D+C-D}{2}\right\}cos\left\{\frac{C+D-C+D}{2}\right\}}{2sin\left\{\frac{C+D-C+D}{2}\right\}cos\left\{\frac{C+D+C-D}{2}\right\}}$
$\Rightarrow \frac{1}{tanAtanB}=\frac{sinCcosD}{cosCsinD}\Rightarrow \frac{-1}{tanAtanB}=\frac{tanC}{tanD}\Rightarrow -tanD=tanAtanB+tanC=0\Rightarrow -tanA=tanBtanC=-tanD\Rightarrow -tanA=tanBtanC+tanD=0$