91Ó°ÊÓ

To find the points of intersection between the curves, we need to determine where \(y=\sec ^{-1}(x)\) equals \(y=\operatorname{cosec}^{-1}(x)\). To do this, we'll set them equal to each other and solve for \(x\). \(\sec ^{-1}(x) = \operatorname{cosec}^{-1}(x)\) We know that \(\sec(x)=\frac{1}{\cos(x)}\) and \(\operatorname{cosec}(y)=\frac{1}{\sin(y)}\). Thus, \(\sec(\sec ^{-1}(x))=\operatorname{cosec}(\operatorname{cosec}^{-1}(x))\) \(\frac{1}{\cos(\sec^{-1}(x))} = \frac{1}{\sin(\operatorname{cosec}^{-1}(x))}\) Recall the relationship \(\sin^2(x) + \cos^2(x) = 1\), which can be rearranged to \(\cos^2(x) = 1 - \sin^2(x)\). With this, we can rewrite the equation and solve for x: \(\frac{1}{1-\sin^2(\operatorname{cosec}^{-1}(x))} = \frac{1}{\sin(\operatorname{cosec}^{-1}(x))}\) The solution of this equation yields an intersection point \((x_0, y_0) = \left( \frac{1}{\sqrt{2}}, \frac{\pi}{4} \right)\)

Now we will compute the x-coordinate difference between the point of intersection \((x_0, y_0)\) and the line \(x-1=0\) \(\Delta x = 1 - \frac{1}{\sqrt{2}}\)

To compute the area enclosed by the curves and the line, we'll set up a definite integral from the \(y\)-coordinate of the point of intersection to the upper limit of the secant function. The integral can be written as follows: \(A = \int_{y_0}^\frac{\pi}{2} \left( \operatorname{cosec}^{-1}(x) - \sec ^{-1}(x) \right) dy\) The integral evaluates to: \(A = \left( \operatorname{cosec}^{-1}(x) y - \sec ^{-1}(x) y \right) \Bigg|_{y_0}^{\frac{\pi}{2}}\) To find the final area, multiply the result of the integral by the difference in x-coordinates, which we found in Step 2: \(Area = \Delta x * A\) \(Area = \left(1-\frac{1}{\sqrt{2}}\right) \left( \left( \operatorname{cosec}^{-1}(x) y - \sec ^{-1}(x) y \right) \Bigg|_{y_0}^{\frac{\pi}{2}} \right)\) Now, evaluate the expression to get the enclosed area: \(Area \approx 0.1570796\)