Prove that tan⁡70∘=tan⁡20∘+2tan⁡50∘ My approach: LHS=tan⁡70∘=1cot⁡70∘=1tan⁡20∘ RHS=tan⁡20∘+2tan⁡50∘ =tan⁡20∘+2tan⁡(20+30)∘ =tan⁡20∘+2(tan⁡20∘+13)1−tan⁡20∘3 =2+33tan⁡20∘−tan220∘3−tan⁡20∘ Why are the two sides not equal despite being expressed in the same terms?

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Prove that tan⁡70∘=tan⁡20∘+2tan⁡50∘  My approach:  LHS=tan⁡70∘=1cot⁡70∘=1tan⁡20∘  RHS=tan⁡20∘+2tan⁡50∘   =tan⁡20∘+2tan⁡(20+30)∘   =tan⁡20∘+2(tan⁡20∘+13)1−tan⁡20∘3   =2+33tan⁡20∘−tan220∘3−tan⁡20∘  Why are the two sides not equal despite being expressed in the same terms?

David Clayton · Accepted Answer

You can do the following: tan⁡(50)=tan⁡(70−20)=tan⁡(70)−tan⁡(20)1+tan⁡(70)tan⁡(20) The bottom is 2, so you are done.  In general, if A+B=π2   then   tan⁡A=tan⁡B+2tan⁡(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∘+2tan⁡50∘  =tan⁡20∘+2tan⁡(70−20)∘  =tan⁡20∘+2(tan⁡70∘−tan⁡20∘)1+tan⁡70∘tan⁡20∘=tan⁡20∘+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.

Lindsey Gamble · Accepted Answer

This will answer the question you asked in bold.  In fact, the two expressions you got involving tan⁡20∘ are equal, not in the sense that algebraic manipulation will produce identical expressions, but rather depending upon a special fact regarding tan⁡20∘. You want to show for x=tan⁡20∘,  1x=2+33x−x23−x  ⇔x3−33x2−3x+3=0  To do this, invoke the fact that 3×20=60 and the formula for tan⁡3x:  3=3x−x31−3x2  and this rearranges to the same cubic equation above.

Vasquez · Accepted Answer

You can use the identity 2cot⁡2A=cot⁡A−tan⁡A which makes the job easy. tan⁡20∘+2tan⁡50∘=tan⁡20∘+2cot⁡40∘ =tan⁡20∘+cot⁡20∘−tan⁡20∘=cot⁡20∘=tan⁡70∘ Otherwise proceed from tan⁡20∘+2tan⁡50∘=tan⁡20∘+2cot⁡40∘ as below - tan⁡20∘+2cot⁡40∘=tan⁡20∘+2cos2⁡20∘−sin2⁡20∘2cos20∘sin⁡20∘ =tan⁡20∘+cot⁡20∘−tan⁡20∘=tan⁡70∘

Prove that \tan 70^{\circ}=\tan 20^{\circ}+2 \tan 50^{\circ} My approach: LHS=\tan 70^{\circ}=\frac{1}{\cot 70^{\circ}}=\frac{1}{\tan

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