MCQ
${d \over {dx}}\{ \log (\sec x + \tan x)\} = $
- A$\cos x$
- ✓$\sec x$
- C$\tan x$
- D$\cot x$
✓
Answer
Correct option: B.
$\sec x$
b
(b) $\frac{d}{{dx}}\{ \log (\sec x + \tan x)\} = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}} = \sec x$.
(b) $\frac{d}{{dx}}\{ \log (\sec x + \tan x)\} = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}} = \sec x$.
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