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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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$7 / 5$
Given the curves $y=e^{2(1+x)-4}$ and $x^2 y=1$
Now $y=e^{2(1+x)-4} \Rightarrow y^{\prime}=2 e^{2 x-2}$
$y^{\prime}(1)=2=m_1$
and $x^2 y=1 \Rightarrow y=\frac{1}{x^2}$
$\Rightarrow y^{\prime}=\frac{-2}{x^3} \Rightarrow y^{\prime}(1)=+2=m_2$
Since, $\tan \theta=\left|\frac{2-(-2)}{1+2(-2)}\right| \Rightarrow \tan \theta=\frac{4}{3}$
So $|\sin \theta|+|\cos \theta|=\frac{4}{5}+\frac{3}{5}=\frac{7}{5}$
Now $y=e^{2(1+x)-4} \Rightarrow y^{\prime}=2 e^{2 x-2}$
$y^{\prime}(1)=2=m_1$
and $x^2 y=1 \Rightarrow y=\frac{1}{x^2}$
$\Rightarrow y^{\prime}=\frac{-2}{x^3} \Rightarrow y^{\prime}(1)=+2=m_2$
Since, $\tan \theta=\left|\frac{2-(-2)}{1+2(-2)}\right| \Rightarrow \tan \theta=\frac{4}{3}$
So $|\sin \theta|+|\cos \theta|=\frac{4}{5}+\frac{3}{5}=\frac{7}{5}$
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