Question

If $\tan {70}^{\circ }=\tan {20}^{\circ }+2\tan A^{\circ }$,then the value of $A$ is (in degree)

Answer 1

$\tan {70}^{\circ }=\tan ({20}^{\circ }+{50}^{\circ })\Rightarrow \tan {70}^{\circ }=\frac{\tan {20}^{\circ }+\tan {50}^{\circ }}{1-\tan {20}^{\circ }\tan {50}^{\circ }}\Rightarrow \tan {70}^{\circ }-\tan {70}^{\circ }\tan {20}^{\circ }\tan {50}^{\circ }=\tan {20}^{\circ }+\tan {50}^{\circ }\Rightarrow \tan {70}^{\circ }-\tan ({90}^{\circ }-{20}^{\circ })\tan {20}^{\circ }\tan {50}^{\circ }=\tan {20}^{\circ }+\tan {50}^{\circ }\Rightarrow \tan {70}^{\circ }-\cot {20}^{\circ }\tan {20}^{\circ }\tan {50}^{\circ }=\tan {20}^{\circ }+\tan {50}^{\circ }\Rightarrow \tan {70}^{\circ }-\tan {50}^{\circ }=\tan {20}^{\circ }+\tan {50}^{\circ }\Rightarrow \tan {70}^{\circ }=\tan {20}^{\circ }+2\tan {50}^{\circ }\therefore A={50}^{\circ }$