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(A) The curves are $y = a^x$ and $y = b^x$. They intersect at the point where $a^x = b^x$,which implies $x = 0$. At $x = 0$,$y = a^0 = 1$. So,the poin

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$\frac{\log a - \log b}{1 + \log a \log b}$

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(A) The curves are $y = a^x$ and $y = b^x$. They intersect at the point where $a^x = b^x$,which implies $x = 0$. At $x = 0$,$y = a^0 = 1$. So,the point of intersection is $(0, 1)$.Now,find the slopes of the tangents at $(0, 1)$:For $y = a^x$,$\frac{dy}{dx} = a^x \log a$. At $x = 0$,$m_1 = a^0 \log a = \log a$.For $y = b^x$,$\frac{dy}{dx} = b^x \log b$. At $x = 0$,$m_2 = b^0 \log b = \log b$.The angle $\alpha$ between the curves is given by $	an \alpha = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.Substituting the values,we get $	an \alpha = |\frac{\log a - \log b}{1 + \log a \log b}|$.

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+\|x-1\|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

If $\alpha$ is the angle of intersection between the curves $y = a^x$ and $y = b^x$,then what is $\tan \alpha$ equal to?

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