Question

If $\tan {22}^o+\tan {38}^o-\sqrt{3}=k\tan {22}^o\tan {38}^o$ then find $k$

• A. $\sqrt{3}$
  
• B. $-\sqrt{3}$
• C. $\sqrt{7}$
• D. 5‾√5

Answer 1

The correct option is B $-\sqrt{3}$

$\tan ({22}^{\circ }+{38}^{\circ })=\frac{\tan {22}^{\circ }+\tan {38}^{\circ }}{1-\tan {22}^{\circ }\tan {38}^{\circ }}$

$\tan \left({60}^{\circ }\right)=\sqrt{3}=\frac{\tan {22}^{\circ }+\tan {38}^{\circ }}{1-\tan {22}^{\circ }\tan {38}^{\circ }}$

$\sqrt{3}-\sqrt{3}\tan {22}^{\circ }\tan {38}^{\circ }=\tan {22}^{\circ }+\tan {38}^{\circ }$

$\tan {22}^{\circ }+\tan {38}^{\circ }-\sqrt{3}=-\sqrt{3}\tan {22}^{\circ }\tan {38}^{\circ }$

$k=-\sqrt{3}$