Question

Prove that: $\tan {70}^{\circ }=\tan {20}^{\circ }+2\tan {50}^{\circ }$.

Answer 1

Consider $\tan {70}^{\circ }$

$\tan {70}^{\circ }=\tan ({20}^{\circ }+{50}^{\circ })$

According to the trigonometric identity,

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

$\tan {70}^{\circ }=\frac{(\tan {20}^{\circ }+\tan {50}^{\circ })}{1-\tan {20}^{\circ }\tan {50}^{\circ }}$

$\tan {70}^{\circ }-\tan {20}^{\circ }\tan {50}^{\circ }\tan {70}^{\circ }=\tan {20}^{\circ }+\tan {50}^{\circ }$

Also, $\tan {70}^{\circ }\tan {20}^{\circ }=\tan {70}^{\circ }\cot {70}^{\circ }=1$ ........ $\left[\tan \right({90}^{\circ }-B)=\cot B]$

Hence,

$\tan {70}^{\circ }-\tan {50}^{\circ }=\tan {20}^{\circ }+\tan {50}^{\circ }$

So, $\tan {70}^{\circ }=\tan {20}^{\circ }+2\tan {50}^{\circ }$. Hence, proved