Question

The solution of ${\log }_{1/2}\sin \theta >{\log }_{1/2}\cos \theta$ in $[0,2\pi ]$ is

• A. $(0,\frac{\pi }{2})$
• B. $(\frac{\pi }{4},\frac{\pi }{2})$
• C. $(0,\frac{\pi }{4})$
• D. None

Answer 1

Answer: (B)

$(\frac{\pi }{4},\frac{\pi }{2})$

${\log }_{\frac{1}{2}}\sin \theta >{\log }_{\frac{1}{2}}\cos \theta$

Since bases are same, we have

${\log }_{\frac{1}{2}}\sin \theta -{\log }_{\frac{1}{2}}\cos \theta >0$

${\log }_{\frac{1}{2}}\frac{\sin \theta }{\cos \theta }>0$

$\Rightarrow \frac{\sin \theta }{\cos \theta }>{\left(\frac{1}{2}\right)}^0$

$\Rightarrow \tan \theta >1$

$\therefore \theta \in (\frac{\pi }{4},\frac{\pi }{2})$