Showing $\tan70° = \tan20° + 2\tan50°$

Question

Q. Prove that $\tan 70° = \tan 20° + 2\tan 50°$.

My approach:

$ LHS = \tan 70° = \dfrac1{\cot 70°} = \dfrac1{\tan 20°}$

$ \begin{align} RHS &= \tan 20° + 2\tan 50° \\ &= \tan 20° + 2\tan (20+30)° \\ &= \tan 20° + \dfrac{2(\tan 20° + 1/√3)}{1 - \tan 20°/√3} \\ &=\dfrac{2 + 3√3 \tan 20° - \tan^2 20°)}{√3 - \tan 20°} \end{align} $

Why are the two sides not equal despite being expressed in the same terms? Can someone offer some help?

Much to my surprise, my friend just expanded tan70° and cross-multiplied the terms to prove it.

Answer 1

You can do the following: $\tan (50) = \tan (70 - 20)= \frac{\tan (70) - \tan (20)} {1+ \tan( 70). \tan (20)}.$ The bottom is $2,$ so you are done.

In general, if $ A +B = \pi/2,$ then $\tan A = \tan B + 2\tan(A-B)$, where $A$ is the bigger one.

Edit: If you can also start from the right/left side. Essentially do the same trick to get equality. For example, RHS= $\tan 20° + 2\tan 50° \\ =\tan 20° + 2\tan (70-20)° \\ = \tan 20° + \frac{2(\tan 70° - \tan 20° )}{1 + \tan 70° \tan 20°} = \tan 20° + \frac{2(\tan 70° - \tan 20° )}{2}\\ =\tan 20° + \tan 70° - \tan 20° .\\ $

To show this equality, you split the angle differently, and the expression has become a bit complicated-looking and Tavish explained why they have to be equal.

Answer 2

You can use the identity $2 \cot 2 A = \cot A - \tan A$ which makes the job easy.

$\tan 20° + 2\tan 50° = \tan 20° + 2\cot 40°$

$= \tan 20° + \cot 20° - \tan 20^0 = \cot 20^0 = \tan 70^0$

Otherwise proceed from $\tan 20° + 2\tan 50° = \tan 20° + 2\cot 40°$ as below -

$\displaystyle \tan 20^0 + 2 \cot 40^0 = \tan 20° + 2 \cdot \frac{\cos^2 20^0 - \sin^2 20^0}{2 \cos 20^0 \sin 20^0}$

$= \tan 20^0 + \cot 20^0 - \tan 20^0 = \tan 70^0$

Answer 3

This will answer the question you asked in bold.

In fact, the two expressions you got involving $\tan 20^\circ$ are equal, not in the sense that algebraic manipulation will produce identical expressions, but rather depending upon a special fact regarding $\tan 20^\circ$. You want to show for $x=\tan 20^\circ$, $$\frac 1x = \frac{2+3\sqrt 3 x -x^2}{\sqrt 3-x}\\ \iff x^3-3\sqrt 3x^2 -3x +\sqrt 3 =0 $$ To do this, invoke the fact that $3\times 20 =60$ and the formula for $\tan 3x$: $$\sqrt 3 = \frac{3x -x^3}{1-3x^2} $$ and this rearranges to the same cubic equation above.