Question

If $\frac{\tan 2^{\circ }\cdot \tan {2017}^{\circ }\cdot \tan {2019}^{\circ }}{\tan {2019}^{\circ }-\tan {2017}^{\circ }-\tan 2^{\circ }}=a$, then the value of ${\tan }^{-1}a$(in degrees) is

Answer 1

${2019}^{\circ }={2017}^{\circ }+2^{\circ }$
Taking $\tan$ on both sides, we get
$\tan {2019}^{\circ }=\tan ({2017}^{\circ }+2^{\circ })\Rightarrow \tan {2019}^{\circ }=\frac{\tan {2017}^{\circ }+\tan 2^{\circ }}{1-\tan {2019}^{\circ }\tan 2^{\circ }}\Rightarrow \tan {2019}^{\circ }-\tan {2019}^{\circ }\tan {2019}^{\circ }\tan 2^{\circ }=\tan {2017}^{\circ }+\tan 2^{\circ }\Rightarrow \frac{\tan 2^{\circ }\cdot \tan {2017}^{\circ }\cdot \tan {2019}^{\circ }}{\tan {2019}^{\circ }-\tan {2017}^{\circ }-\tan 2^{\circ }}=1\therefore {\tan }^{-1}a={45}^{\circ }$