A typical way of finding an inverse function is to set y = f(x)y=f(x) and then solve for xx to obtain x = f^-1(y)x=f−1(y)
Applying that here, we start with
y = -ln(arctan(x))y=−ln(arctan(x))
=> -y = ln(arctan(x))⇒−y=ln(arctan(x))
=>e^-y = e^(ln(arctan(x))) = arctan(x)⇒e−y=eln(arctan(x))=arctan(x) (by the definition of lnln)
=> tan(e^-y) = tan(arctan(x)) = x⇒tan(e−y)=tan(arctan(x))=x (by the definition of arctanarctan)
Thus we have f^-1(x) = tan(e^-x)f−1(x)=tan(e−x)
If we wish to confirm this via the definition f^-1(f(x)) = f(f^-1(x)) = xf−1(f(x))=f(f−1(x))=x
remember that y = f(x)y=f(x) so we already have
f^-1(y) = f^-1(f(x)) = xf−1(y)=f−1(f(x))=x
For the reverse direction,
f(f^-1(x)) = -ln(arctan(tan(e^-x))f(f−1(x))=−ln(arctan(tan(e−x))
=> f(f^-1(x)) = -ln(e^-x)⇒f(f−1(x))=−ln(e−x)
=> f(f^-1(x)) = -(-x*ln(e)) = -(-x*1)⇒f(f−1(x))=−(−x⋅ln(e))=−(−x⋅1)
=> f(f^-1(x)) = x⇒f(f−1(x))=x
