If \( \cos(A+B)\sin(C-D)=\cos(A-B)\sin(C+D) \), then write the value of \( \tan A\tan B\tan C \)
Solution:
Using identities,
\[
\cos(A+B)=\cos A\cos B-\sin A\sin B
\]
and
\[
\cos(A-B)=\cos A\cos B+\sin A\sin B
\]
Also,
\[
\sin(C-D)=\sin C\cos D-\cos C\sin D
\]
and
\[
\sin(C+D)=\sin C\cos D+\cos C\sin D
\]
Substituting,
\[
(\cos A\cos B-\sin A\sin B)
(\sin C\cos D-\cos C\sin D)
\]
\[
=
(\cos A\cos B+\sin A\sin B)
(\sin C\cos D+\cos C\sin D)
\]
Expanding and cancelling common terms,
\[
\cos A\cos B\cos C\sin D
+
\sin A\sin B\sin C\cos D
=0
\]
\[
\sin A\sin B\sin C\cos D
=
-\cos A\cos B\cos C\sin D
\]
Dividing by
\[
\cos A\cos B\cos C\cos D
\]
\[
\tan A\tan B\tan C
=
-\tan D
\]
\[
\boxed{-\tan D}
\]