Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{\cos \left(A-B\right)}{\cos \left(A+B\right)}+\frac{\cos \left(C+D\right)}{\cos \left(C-D\right)}=0 \\ \Rightarrow \frac{\cos \left(A-B\right)\cos \left(C-D\right)+\cos \left(C+D\right)\cos \left(A+B\right)}{\cos \left(A+B\right)\cos \left(C-D\right)}=0 \\ \Rightarrow \cos \left(A-B\right)\cos \left(C-D\right)+\cos \left(C+D\right)\cos \left(A+B\right)=0 \\ \Rightarrow \cos \left(A-B\right)\cos \left(C-D\right)=-\cos \left(C+D\right)\cos \left(A+B\right) \\ \Rightarrow \left[\cos A\cos B+\sin A\sin B\right]\left[\cos C\cos D+\sin C\sin D\right]=-\left[\cos C\cos D-\sin C\sin D\right]\left[\cos A\cos B-\sin A\sin B\right] \end{gathered}$$

$$\begin{gathered} \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\cos A\cos B\cos C\cos Dweget, \\ \frac{\left[\cos A\cos B+\sin A\sin B\right]\left[\cos C\cos D+\sin C\sin D\right]}{\cos A\cos B\cos C\cos D}=-\frac{\left[\cos C\cos D-\sin C\sin D\right]\left[\cos A\cos B-\sin A\sin B\right]}{\cos A\cos B\cos C\cos D} \\ \Rightarrow \frac{\left[\cos A\cos B+\sin A\sin B\right]}{\cos A\cos B}\times \frac{\left[\cos C\cos D+\sin C\sin D\right]}{\cos C\cos D}=-\frac{\left[\cos C\cos D-\sin C\sin D\right]}{\cos C\cos D}\times \frac{\left[\sin C\cos A\cos B-\sin A\sin B\right]}{\cos A\cos B} \\ \Rightarrow \left[1+\tan A\tan B\right]\left[1+\tan C\tan D\right]=\left[\tan C\tan D-1\right]\left[1-\tan A\tan B\right] \\ \Rightarrow 1+\tan C\tan D+\tan A\tan B+\tan A\tan B\tan C\tan D=\tan C\tan D-\tan A\tan B\tan C\tan D+\tan A\tan B\tan D-1+\tan A\tan B \\ \Rightarrow 2\tan A\tan B\tan C\tan D=-2 \\ \Rightarrow \tan A\tan B\tan C\tan D=-1 \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$