In a $\triangle ABC$, $\angle B < \angle C$ and values of $B$ and $C$ satisfy equation $2 \tan x – k (1 + \tan ^2x) = 0$ where $(0 \lt k \lt 1)$ . Then measure of angle $A$ is?
Note that $\tan B$ and $\tan C$ are its roots , So $$\text{sum of roots} = \dfrac2k$$ and $$\text{product of roots} = 1\,.$$ Then , $$\tan ( B + C) = \tan ( \pi - A)\,$$ Expanding $\tan( B + C)$, we found it not defined. The tangent is not defined at when the argument is $\dfrac{\pi}{2}$. So, $$\tan (\pi - A) = \tan\left(\frac{\pi}{2}\right)\text{ and }\angle A=\frac{\pi}{2} $$