Question

If $\frac{\tan 2^{\circ }\cdot \tan {2017}^{\circ }\cdot \tan {2019}^{\circ }}{\tan {2019}^{\circ }-\tan {2017}^{\circ }-\tan 2^{\circ }}=a$ such that $a=\tan x,$ where $x$ is an acute angle in degree, then the value of $x$ is

• A. $60$
• B. $0$
• C. $15$
• D. $45$

Answer 1

The correct option is D $45$

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

As $\tan {45}^{\circ }=1,$
$\therefore x={45}^{\circ }$