Question

The value of $\alpha$ lying between $[0,\pi ]$ for which the inequality $\tan \alpha >{\tan }^3\alpha$ is valid, is

• A. $(0,\frac{\pi }{4})\cup (\frac{\pi }{2},\frac{3\pi }{4})$
• B. $(0,\frac{\pi }{2})\cup (\frac{\pi }{2},\frac{3\pi }{4})$
• C. $(0,\frac{\pi }{3})\cup (\frac{\pi }{3},\frac{\pi }{2})$
• D. $(0,\frac{\pi }{4})\cup (\frac{\pi }{4},\frac{3\pi }{4})$

Answer 1

The correct option is A $(0,\frac{\pi }{4})\cup (\frac{\pi }{2},\frac{3\pi }{4})$

We have : $\tan \alpha >{\tan }^3\alpha$+
$\tan \alpha -{\tan }^3\alpha >0\Rightarrow \tan \alpha (1-{\tan }^2\alpha )>0$
$\Rightarrow \left(\tan \alpha \right)(\tan \alpha +1)(\tan \alpha -1)<0$
So using wavy curve method $\tan \alpha \in (-\infty ,-1)\cup (0,1)$


$\therefore \alpha \in (0,\frac{\pi }{4})\cup (\frac{\pi }{2},\frac{3\pi }{4})$