Question

Prove that $\tan {70}^o-\tan {20}^o=2\tan {50}^o$

Answer 1

We know that

$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\Rightarrow \tan \left({50}^0\right)=\tan ({70}^0-{20}^0)=\frac{\tan {70}^0-\tan {20}^0}{1+\tan {70}^0\tan {20}^0}$

$\Rightarrow \tan \left({50}^0\right)=\frac{\tan {70}^0-\tan {20}^0}{1+\tan {70}^0\tan {20}^0}$ .......$\left(1\right)$

$\Rightarrow \tan \left({90}^0\right)=\tan ({70}^0+{20}^0)=\frac{\tan {70}^0+\tan {20}^0}{1-\tan {70}^0\tan {20}^0}$

$\Rightarrow \tan \left({90}^0\right)=\frac{\tan {70}^0+\tan {20}^0}{1-\tan {70}^0\tan {20}^0}$

Therefore, $\Rightarrow 1-\tan {70}^0\tan {20}^0=0$

$\Rightarrow \tan {70}^0\tan {20}^0=1$

Substitute this result in $\left(1\right)$ we get

$\Rightarrow \tan \left({50}^0\right)=\frac{\tan {70}^0-\tan {20}^0}{1+1}$

$\Rightarrow \tan \left({50}^0\right)=\frac{\tan {70}^0-\tan {20}^0}{2}$

$\Rightarrow \tan {70}^0-\tan {20}^0=2\tan {50}^0$

Hence proved.