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Let $y=e^{x^{2}}$ and $y=e^{x^{2}} \sin x$ be two given curves. Then, angle between the tangents to the curves at any point of their intersection
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For intersecting points, $e^{x^{2}}=e^{x^{2}} \sin x$ $\Rightarrow \quad e^{x^{2}}(\sin x-1)=0$
$\Rightarrow \quad e^{x^{2}}=0 \quad$ or $\sin x=1$
But $\quad e^{x^{2}} \neq 0 \Rightarrow \sin x=1$
$\Rightarrow$
$$
x=\frac{\pi}{2}
$$
Now.
$$
y=e^{x^{2}}
$$
$\therefore \quad \frac{d y}{d x}=e^{x^{2}} \cdot 2 x=2 x e^{x^{2}}$
Also, $y=e^{x^{2}} \sin x$
$\frac{d y}{d x}=e^{x^{2}} \cdot 2 x \sin x+e^{x^{2}} \cos x=2 x e^{x^{2}}$
$[\because \sin x=1, \cos x=0]$
Since, both the curves has equal slope. Hence, angle between the tangents at intersecting point is $0 .$
$\Rightarrow \quad e^{x^{2}}=0 \quad$ or $\sin x=1$
But $\quad e^{x^{2}} \neq 0 \Rightarrow \sin x=1$
$\Rightarrow$
$$
x=\frac{\pi}{2}
$$
Now.
$$
y=e^{x^{2}}
$$
$\therefore \quad \frac{d y}{d x}=e^{x^{2}} \cdot 2 x=2 x e^{x^{2}}$
Also, $y=e^{x^{2}} \sin x$
$\frac{d y}{d x}=e^{x^{2}} \cdot 2 x \sin x+e^{x^{2}} \cos x=2 x e^{x^{2}}$
$[\because \sin x=1, \cos x=0]$
Since, both the curves has equal slope. Hence, angle between the tangents at intersecting point is $0 .$
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