If \[ \cos(A+B)\sin(C-D) = \cos(A-B)\sin(C+D) \] prove that \[ \tan A\tan B\tan C+\tan D=0 \]
Solution
Given:
\[ \cos(A+B)\sin(C-D) = \cos(A-B)\sin(C+D) \]Use identities:
\[ \cos(X+Y)=\cos X\cos Y-\sin X\sin Y \] \[ \cos(X-Y)=\cos X\cos Y+\sin X\sin Y \] \[ \sin(X+D)=\sin X\cos D+\cos X\sin D \] \[ \sin(X-D)=\sin X\cos D-\cos X\sin D \]
\[
(\cos A\cos B-\sin A\sin B)
(\sin C\cos D-\cos C\sin D)
\]
\[
=
(\cos A\cos B+\sin A\sin B)
(\sin C\cos D+\cos C\sin D)
\]
Expand both sides:
\[ \cos A\cos B\sin C\cos D -\cos A\cos B\cos C\sin D \] \[ -\sin A\sin B\sin C\cos D +\sin A\sin B\cos C\sin D \] \[ = \cos A\cos B\sin C\cos D +\cos A\cos B\cos C\sin D \] \[ +\sin A\sin B\sin C\cos D +\sin A\sin B\cos C\sin D \]Cancel common terms and simplify:
\[ -2\cos A\cos B\cos C\sin D = 2\sin A\sin B\sin C\cos D \]
\[
-\cos A\cos B\cos C\sin D
=
\sin A\sin B\sin C\cos D
\]
Divide by
\[ \cos A\cos B\cos C\cos D \]
\[
-\tan D
=
\tan A\tan B\tan C
\]
\[
\tan A\tan B\tan C+\tan D=0
\]
Hence Proved.