Find the Value of tan 20° + tan 40° + √3 tan 20° tan 40°

Solution

We use the identity:

\[ \tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B} \]

Taking

\[ A=20^\circ, \qquad B=40^\circ \]

Then,

\[ \tan60^\circ = \frac{\tan20^\circ+\tan40^\circ} {1-\tan20^\circ\tan40^\circ} \]

Since

\[ \tan60^\circ=\sqrt{3} \]

we get

\[ \sqrt{3} = \frac{\tan20^\circ+\tan40^\circ} {1-\tan20^\circ\tan40^\circ} \]

Cross multiplying,

\[ \sqrt{3}(1-\tan20^\circ\tan40^\circ) = \tan20^\circ+\tan40^\circ \]

Expanding,

\[ \sqrt{3} – \sqrt{3}\tan20^\circ\tan40^\circ = \tan20^\circ+\tan40^\circ \]

Bringing all terms to one side,

\[ \tan20^\circ+\tan40^\circ + \sqrt{3}\tan20^\circ\tan40^\circ = \sqrt{3} \]

Final Answer

\[ \boxed{ \tan20^\circ+\tan40^\circ+\sqrt{3}\tan20^\circ\tan40^\circ = \sqrt{3} } \]

Correct Option: (c)