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



Answered question


Prove that tan70=tan20+2tan50
My approach:
Why are the two sides not equal despite being expressed in the same terms?

Answer & Explanation

David Clayton

David Clayton

Beginner2021-12-31Added 36 answers

You can do the following: tan(50)=tan(7020)=tan(70)tan(20)1+tan(70)tan(20) The bottom is 2, so you are done.
In general, if A+B=π2   then   tanA=tanB+2tan(AB) 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=tan20+2tan50
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

Lindsey Gamble

Beginner2022-01-01Added 38 answers

This will answer the question you asked in bold.
In fact, the two expressions you got involving tan20 are equal, not in the sense that algebraic manipulation will produce identical expressions, but rather depending upon a special fact regarding tan20. You want to show for x=tan20,
To do this, invoke the fact that 3×20=60 and the formula for tan3x:
and this rearranges to the same cubic equation above.


Expert2022-01-08Added 669 answers

You can use the identity 2cot2A=cotAtanA which makes the job easy.
Otherwise proceed from tan20+2tan50=tan20+2cot40 as below -

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