Explanation

(c)
$When\ x = \frac{\pi}{4}$
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$y = \tan{\frac{\pi}{4} = 1}$
$\frac{dy}{dx} = \sec^{2}x\ \ \ \ \ \ \lbrack\because y = \tan{x\rbrack}$
$\Longrightarrow \ \ \left\lbrack \frac{dy}{dx} \right\rbrack_{x = \pi/4} = 2$
$Equation\ of\ tangent\ at\ P\left( \frac{\pi}{4},1 \right)is$
$y - 1 = 2\left( x - \frac{\pi}{4} \right) \Longrightarrow y = 2x + 1 - \frac{\pi}{2}$
$It\ meets\ x–axis\ at\ T\left( \frac{\pi - 2}{4},0 \right)$
$Required\ area = \int_{0}^{\pi/4}{\tan{x\ dx - \frac{1}{2}TN \bullet PN}}$
$= \left\lbrack \log{\sec x} \right\rbrack_{0}^{\pi/4} - \frac{1}{2} \bullet \frac{1}{2} \bullet 1$
$\left\lbrack \because TN = ON - OT = \frac{\pi}{4} - \frac{\pi - 2}{4} = \frac{1}{2} \right\rbrack$
$= \log{\sqrt{2} - 0 - \frac{1}{4} = \left( \log\sqrt{2} - \frac{1}{4} \right)sq\ unit}$