Q56 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics
For the value of $\frac{2 \tan (x)}{1-\tan ^2(x)}$ to be positive, find values of $x$, such that $x \in\left(0, \frac{\pi}{2}\right)$
- A$\left(0, \frac{\pi}{3}\right)$
- B$\left(0, \frac{\pi}{6}\right)$
- Correct answerC$\left(0, \frac{\pi}{4}\right)$
- D$\left(0, \frac{\pi}{8}\right)$
Solution
Given,
$$
\begin{aligned}
& \frac{2 \tan x}{1-\tan ^2 x}>0 \\
& \tan 2 x>0
\end{aligned}
$$
Since, $\quad 0 < 2 x < \frac{\pi}{2} \quad\left[\because \tan 2 x\right.$ is the in $\left.Q_1\right]$
$$
\begin{array}{ll}
\Rightarrow & 0 < x < \frac{\pi}{4} \\
\therefore & 0 < x < \frac{\pi}{4} \text { as } x \in\left(0, \frac{\pi}{2}\right)
\end{array}
$$
Hence, option (3) is correct.
Hence, the correct answer is option: 'C'.