Question

If $y=\log (\sec x+\tan x),$ then $\frac{dy}{dx}=$

• A. $\tan x$
• B. $secx$
• C. ${\tan }^{2}x$
• D. none of these

Answer 1

Answer: (B)

$\frac{dy}{dx}=\frac{1}{\sec x+\tan x}.\left(\sec x\tan x+{\sec }^{2}x\right)$
$=\frac{\sec x\left[\tan x+\sec x\right]}{\sec x+\tan x}=\sec x$