kramtus51

2021-12-30

Prove that ${\mathrm{tan}70}^{\circ }={\mathrm{tan}20}^{\circ }+2{\mathrm{tan}50}^{\circ }$
My approach:
$LHS={\mathrm{tan}70}^{\circ }=\frac{1}{{\mathrm{cot}70}^{\circ }}=\frac{1}{{\mathrm{tan}20}^{\circ }}$
$RHS={\mathrm{tan}20}^{\circ }+2{\mathrm{tan}50}^{\circ }$
$={\mathrm{tan}20}^{\circ }+2{\mathrm{tan}\left(20+30\right)}^{\circ }$
$={\mathrm{tan}20}^{\circ }+\frac{2\left({\mathrm{tan}20}^{\circ }+\frac{1}{\sqrt{3}}\right)}{1-\frac{{\mathrm{tan}20}^{\circ }}{\sqrt{3}}}$
$=\frac{2+3\sqrt{3}{\mathrm{tan}20}^{\circ }-{\mathrm{tan}}^{2}{20}^{\circ }}{\sqrt{3}-{\mathrm{tan}20}^{\circ }}$
Why are the two sides not equal despite being expressed in the same terms?

David Clayton

You can do the following: $\mathrm{tan}\left(50\right)=\mathrm{tan}\left(70-20\right)=\frac{\mathrm{tan}\left(70\right)-\mathrm{tan}\left(20\right)}{1+\mathrm{tan}\left(70\right)\mathrm{tan}\left(20\right)}$ The bottom is 2, so you are done.
In general, if 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 $RHS={\mathrm{tan}20}^{\circ }+2{\mathrm{tan}50}^{\circ }$
$={\mathrm{tan}20}^{\circ }+2{\mathrm{tan}\left(70-20\right)}^{\circ }$
$={\mathrm{tan}20}^{\circ }+\frac{2\left({\mathrm{tan}70}^{\circ }-{\mathrm{tan}20}^{\circ }\right)}{1+{\mathrm{tan}70}^{\circ }{\mathrm{tan}20}^{\circ }}={\mathrm{tan}20}^{\circ }+\frac{2\left({\mathrm{tan}70}^{\circ }-{\mathrm{tan}20}^{\circ }\right)}{2}$
$={\mathrm{tan}20}^{\circ }+{\mathrm{tan}70}^{\circ }-{\mathrm{tan}20}^{\circ }$
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.

Lindsey Gamble

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

Vasquez

You can use the identity $2\mathrm{cot}2A=\mathrm{cot}A-\mathrm{tan}A$ which makes the job easy.
$\mathrm{tan}{20}^{\circ }+2\mathrm{tan}{50}^{\circ }=\mathrm{tan}{20}^{\circ }+2\mathrm{cot}{40}^{\circ }$
$=\mathrm{tan}{20}^{\circ }+\mathrm{cot}{20}^{\circ }-\mathrm{tan}{20}^{\circ }=\mathrm{cot}{20}^{\circ }=\mathrm{tan}{70}^{\circ }$
Otherwise proceed from $\mathrm{tan}{20}^{\circ }+2\mathrm{tan}{50}^{\circ }=\mathrm{tan}{20}^{\circ }+2\mathrm{cot}{40}^{\circ }$ as below -
$\mathrm{tan}{20}^{\circ }+2\mathrm{cot}{40}^{\circ }=\mathrm{tan}{20}^{\circ }+2\frac{{\mathrm{cos}}^{2}{20}^{\circ }-{\mathrm{sin}}^{2}{20}^{\circ }}{2cos{20}^{\circ }\mathrm{sin}{20}^{\circ }}$
$=\mathrm{tan}{20}^{\circ }+\mathrm{cot}{20}^{\circ }-\mathrm{tan}{20}^{\circ }=\mathrm{tan}{70}^{\circ }$

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