Question

$f\left(x\right)=\frac{1}{2}$ sinx tanx- $\log$ secx is increasing in

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

Answer 1

Answer: (C)

$(0,\frac{\pi }{2})$

$f^{\prime }\left(x\right)=\frac{1}{2}tanxcosx+\frac{1}{2}sinx{\sec }^{2}x-\frac{secx}{secx}tanx$
$=\frac{sinx}{2}+\frac{tanxsecx}{2}-tanx$
$=\frac{1}{2}(sinx+tanx(secx-1\left)\right)$
$f^{\prime }\left(x\right)>0$ If $xϵ(0,\frac{\pi }{2})$