The area of the region ( in square units ) above the x -axis...

The given curve is $y = \tan x$ and at $x = \dfrac{\pi}{4}, y = \tan \left(\dfrac{\pi}{4}\right) = 1$

Also, $\dfrac{dy}{dx} = se{c}^{2}x$ and ${\left(\dfrac{dy}{dx}\right)}_{x = \dfrac{\pi}{4}} = se{c}^{2}\left(\dfrac{\pi}{4}\right) = {\left(\sqrt{2}\right)}^{2} = 2$

We know that the equation of the tangent to a curve $y = f\left(x\right)$ at a point \left({x}_{1}, {y}_{1}\right)$is$y - {y}_{1} = {\left(\dfrac{dy}{dx}\right)}_{x = \dfrac{\pi}{4}}\left(x - {x}_{1}\right)$</p><p>Hence, the equation of the tangent to$y = \tan x$at$P\left(\dfrac{\pi}{4}, 1\right)$is$y - 1 = 2\left(x - \dfrac{\pi}{4}\right)$</p><p>$⇒ y - 1 = 2x - \dfrac{\pi}{2}$</p><p>$⇒ y - 1 + \dfrac{\pi}{2} = 2x$</p><p>To find the point where this line cuts the$x$-axis, put$y = 0,$to get</p><p>$2x = \dfrac{\pi}{2} - 1$</p><p>$⇒ x = \dfrac{\pi}{4} - \dfrac{1}{2}$</p><p>Thus, the point$A ≡ \left(\dfrac{\pi}{4} - \dfrac{1}{2}, 0\right) .$</p><p>Now, the graph for the given information is</p><p><img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/73283292-fd4f-4e8d-ac0f-9933576f6d5d-Untitled-tanx.png" style="width: 221px; height: 250px;" /></p><p>And, the required Area<br /><br />$= {∫}_{0}^{\dfrac{\pi}{4}}\left(\tan x\right)dx - ar\left(∆ PAB\right)$</p><p>$= {\left[\log \left(\sec x\right)\right]}_{0}^{\dfrac{\pi}{4}} - \dfrac{1}{2} \times \left(PB \times AB\right)$<br /><br />$= \left[\log \left(sec\dfrac{\pi}{4}\right) - \log \left(\sec 0\right)\right] - \dfrac{1}{2} \times 1 \times \left(\dfrac{\pi}{4} - \left(\dfrac{\pi}{4} - \dfrac{1}{2}\right)\right)$</p><p>$= \left|\log \left(\sqrt{2}\right) - log\left(1\right)\right| - \dfrac{1}{4}

$= \left|\log \left(2\right)\dfrac{1}{2} - 0\right| - \dfrac{1}{4}$

$= \dfrac{1}{2}\log 2 - \dfrac{1}{4} = \dfrac{1}{2}\left(\log 2 - \dfrac{1}{2}\right)$ sq units.