If the angle between the curves y=e^{2(1+x)-4 | vedclass

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(A) Given the curves $y=e^{2(1+x)-4}$ and $x^2 y=1$. At the point $(1,1)$,we find the slopes of the tangents to both curves. For the first curve $y=e^

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(A) Given the curves $y=e^{2(1+x)-4}$ and $x^2 y=1$. At the point $(1,1)$,we find the slopes of the tangents to both curves. For the first curve $y=e^{2(1+x)-4}$,differentiating with respect to $x$ gives $y' = e^{2(1+x)-4} \cdot 2$. At $(1,1)$,the slope $m_1 = y'(1) = e^{2(1+1)-4} \cdot 2 = e^0 \cdot 2 = 2$. For the second curve $x^2 y = 1$,we have $y = x^{-2}$. Differentiating with respect to $x$ gives $y' = -2x^{-3} = -\frac{2}{x^3}$. At $(1,1)$,the slope $m_2 = y'(1) = -\frac{2}{1^3} = -2$. The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. Substituting the values,$	an 	heta = \left| \frac{2 - (-2)}{1 + (2)(-2)} ight| = \left| \frac{4}{1 - 4} ight| = \left| \frac{4}{-3} ight| = \frac{4}{3}$. Since $	an 	heta = \frac{4}{3}$,we can consider a right-angled triangle with opposite side $4$ and adjacent side $3$. The hypotenuse is $\sqrt{4^2 + 3^2} = 5$. Thus,$\sin 	heta = \frac{4}{5}$ and $\cos 	heta = \frac{3}{5}$. Therefore,$|\sin 	heta| + |\cos 	heta| = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}$.

If the angle between the curves $y=e^{2(1+x)-4}$ and $x^2 y=1$ at the point $(1,1)$ is $\theta$,then $|\sin \theta|+|\cos \theta|=$

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