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$\frac{d}{dx} (\log | \sec x + Tan x |) =__________$

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Sin x

Cos x

Sec x

Sec x Tan x

answer is C.

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Detailed Solution

$\frac{d}{dx} (\log | Secx + Tanx |) = \frac{1}{Secx + Tanx} (Secxtanx + \sec^{2} x)$
$= \frac{Secx (Secx + Tanx)}{Secx + Tanx} = Secx$

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