To prove that $\tan70=\tan20+2\tan50$

Question

to prove that $$\tan70=\tan20+2\tan50$$

Now i wrote $70=20+50$ and took $\tan$ on both sides and after cross multiplying i have got to $\tan70 -\tan20\tan50\tan70=\tan20+\tan50$. how do i proced?

thanks

Answer 1

\begin{align*} \tan70 - \tan 20 & =\dfrac{\sin 70}{\cos 70}-\dfrac{\sin20}{\cos20}\\ & =\dfrac{\sin70\cos20-\cos70\sin20}{\cos20\cos70}\\ & =\dfrac{\sin(70-20)}{\cos20 \cos 70}\\ & =\dfrac{\sin50}{\cos 20 \cos 70}\\ &=\dfrac{\sin50}{\cos20 .\sin(20)}\\ &=\dfrac{\color{red} {2}\sin50}{\color{red} {2}\sin20.\cos 20 }\\ &=\dfrac{\color{red} {2}\sin50}{\sin40 }\\ &=\dfrac{\color{red} {2}\sin50}{\cos(90-40) }\\ &=\dfrac{\color{red} {2}\sin50}{\cos50}\\ &=2\tan50 \end{align*}

Answer 2

Generalization :

$$\tan(90^\circ-A)-\tan A=\cot A-\tan A=\cdots=2\cot2A=2\tan(90^\circ-2A)$$

Answer 3

i got this . just noticed $\tan70\tan20=1$

Answer 4

We need to prove that $$\tan70^{\circ}-\tan50^{\circ}=\tan20^{\circ}+\tan50^{\circ}$$ or $$\frac{\sin20^{\circ}}{\cos70^{\circ}\cos50^{\circ}}=\frac{\sin70^{\circ}}{\cos20^{\circ}\cos50^{\circ}},$$ which is obvious because $\sin20^{\circ}=\cos70^{\circ}$ and $\sin70^{\circ}=\cos20^{\circ}$