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Answer 1 · 2016-11-08T17:07:50

graph{(abs(x^2-1)-y)(abs(x^2-3)-y)((x-sqrt2)^2+(y-1)^2-0.002)=0 [-1, 4, -0.5, 2]}

#alpha=pi-2arctan2sqrt(2)#

The curves can intersect only when the arguments signs are opposite so

#x^2-1=3-x^2\ \ \ \ =>\ \ \ \ x=+-sqrt2#

#x_0=sqrt(2)#

#f_(1,2)^'(x_0)=+-2sqrt2#

So the obtuse angle is

#beta=arctan2sqrt2-arctan(-2sqrt2)=2arctan2sqrt2#

then

#alpha=pi-2arctan2sqrt(2)#

Now try to convert it into a single arctan value.

#tan(alpha)=-tan(2arctan2sqrt(2))#

Now remember duplication formula for tan.

#tan(2x)=(2tanx)/(1-tan^2x)#

#tan(alpha)=-(2tan(arctan2sqrt(2)))/(1-tan^2(arctan2sqrt2))=-(2*2sqrt2)/(1-4*2)=(4sqrt2)/7#

So

#alpha=arctan((4sqrt2)/7)#