Prove that: \( \tan20^\circ \tan30^\circ \tan40^\circ \tan80^\circ = 1 \)
Solution:
\[
\tan20^\circ \tan30^\circ \tan40^\circ \tan80^\circ
\]
Using identity,
\[
\tan\theta \tan(60^\circ-\theta)\tan(60^\circ+\theta)=\tan3\theta
\]
Putting \( \theta=20^\circ \),
\[
\tan20^\circ \tan40^\circ \tan80^\circ
=
\tan60^\circ
=
\sqrt3
\]
Therefore,
\[
\tan20^\circ \tan30^\circ \tan40^\circ \tan80^\circ
=
\sqrt3 \times \frac{1}{\sqrt3}
\]
\[
=1
\]
\[
\boxed{\tan20^\circ \tan30^\circ \tan40^\circ \tan80^\circ=1}
\]