If \[ \frac{\cos(A-B)}{\cos(A+B)} + \frac{\cos(C+D)}{\cos(C-D)} =0 \] prove that \[ \tan A\tan B\tan C\tan D=-1 \]
Solution
Given:
\[ \frac{\cos(A-B)}{\cos(A+B)} + \frac{\cos(C+D)}{\cos(C-D)} =0 \]Transpose second term:
\[ \frac{\cos(A-B)}{\cos(A+B)} = -\frac{\cos(C+D)}{\cos(C-D)} \]Cross multiply:
\[ \cos(A-B)\cos(C-D) = -\cos(A+B)\cos(C+D) \]Use identity:
\[ \cos(X-Y)=\cos X\cos Y+\sin X\sin Y \] \[ \cos(X+Y)=\cos X\cos Y-\sin X\sin Y \]
\[
(\cos A\cos B+\sin A\sin B)
(\cos C\cos D+\sin C\sin D)
\]
\[
=
-(\cos A\cos B-\sin A\sin B)
(\cos C\cos D-\sin C\sin D)
\]
Expand both sides:
\[ \cos A\cos B\cos C\cos D +\cos A\cos B\sin C\sin D \] \[ +\sin A\sin B\cos C\cos D +\sin A\sin B\sin C\sin D \] \[ = -\cos A\cos B\cos C\cos D +\cos A\cos B\sin C\sin D \] \[ +\sin A\sin B\cos C\cos D -\sin A\sin B\sin C\sin D \]Bring like terms together:
\[ 2\cos A\cos B\cos C\cos D = -2\sin A\sin B\sin C\sin D \]
\[
\cos A\cos B\cos C\cos D
=
-\sin A\sin B\sin C\sin D
\]
Divide by \(\cos A\cos B\cos C\cos D\):
\[ 1 = -\tan A\tan B\tan C\tan D \]
\[
\tan A\tan B\tan C\tan D
=
-1
\]
Hence Proved.