Trigonometry – II — Maths STD 11 Science — Question

\begin{aligned}
& \text { L.H.S. }=\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ} \\
& =\tan 20^{\circ} \tan 80^{\circ} \cot \left(90^{\circ}-40^{\circ}\right) \\
& =\tan 20^{\circ} \tan 80^{\circ} \tan 40^{\circ} \\
& =\tan 20^{\circ} \tan \left(60^{\circ}+20^{\circ}\right) \tan \left(60^{\circ}-20^{\circ}\right) \\
& =\tan 20^{\circ}\left(\frac{\tan 60^{\circ}+\tan 20^{\circ}}{1-\tan 60^{\circ} \tan 20^{\circ}}\right)\left(\frac{\tan 60^{\circ}-\tan 20^{\circ}}{1+\tan 60^{\circ} \tan 20^{\circ}}\right) \\
& =\tan 20^{\circ}\left(\frac{\sqrt{3}+\tan 20^{\circ}}{1-\sqrt{3} \tan 20^{\circ}}\right)\left(\frac{\sqrt{3}-\tan 20^{\circ}}{1+\sqrt{3} \tan 20^{\circ}}\right) \\
& =\tan 20^{\circ}\left[\frac{(\sqrt{3})^2-\tan { }^2 20^{\circ}}{1^2-\left(\sqrt{3} \tan 20^{\circ}\right)^2}\right] \\
& =\tan 20^{\circ}\left(\frac{3-\tan ^2 20^{\circ}}{1-3 \tan ^2 20^{\circ}}\right) \\
& =\frac{3 \tan 20^{\circ}-\tan ^3 20^{\circ}}{1-3 \tan ^2 20^{\circ}} \\
& =\tan 3\left(20^{\circ}\right) \\
& =\tan 60^{\circ} \\
& =\sqrt{3} \\
& =\text { R. H.S. }
\end{aligned}