The area of the region bounded by the curve y | vedclass

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(D) The curve is $y = 	an x$. At $x = \frac{\pi}{4}$,$y = 	an(\frac{\pi}{4}) = 1$. The point of tangency is $(\frac{\pi}{4}, 1)$.The derivative is $

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(D) The curve is $y = 	an x$. At $x = \frac{\pi}{4}$,$y = 	an(\frac{\pi}{4}) = 1$. The point of tangency is $(\frac{\pi}{4}, 1)$.The derivative is $\frac{dy}{dx} = \sec^2 x$. At $x = \frac{\pi}{4}$,the slope $m = \sec^2(\frac{\pi}{4}) = 2$.The equation of the tangent line is $y - 1 = 2(x - \frac{\pi}{4})$,which simplifies to $y = 2x - \frac{\pi}{2} + 1$.The tangent intersects the $x$-axis where $y = 0$,so $0 = 2x - \frac{\pi}{2} + 1$,which gives $x = \frac{\pi}{4} - \frac{1}{2}$.The area is given by the integral $\int_{\frac{\pi}{4} - \frac{1}{2}}^{\frac{\pi}{4}} 	an x \, dx - 	ext{Area of the triangle formed by the tangent}$.Alternatively,integrating with respect to $y$: The curve is $x = \arctan y$. The tangent is $x = \frac{y + \frac{\pi}{2} - 1}{2} = \frac{y}{2} + \frac{\pi}{4} - \frac{1}{2}$.Area $= \int_{0}^{1} (\arctan y - (\frac{y}{2} + \frac{\pi}{4} - \frac{1}{2})) \, dy = [y \arctan y - \frac{1}{2} \log(1 + y^2) - \frac{y^2}{4} - \frac{\pi y}{4} + \frac{y}{2}]_{0}^{1}$.$= (1 \cdot \frac{\pi}{4} - \frac{1}{2} \log 2 - \frac{1}{4} - \frac{\pi}{4} + \frac{1}{2}) - (0) = \frac{1}{4} - \frac{1}{2} \log 2 = \frac{1}{4} - \log \sqrt{2}$.

The area of the region bounded by the curve $y = \tan x$,the tangent drawn to the curve at $x = \frac{\pi}{4}$,and the $x$-axis is:

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