In the book:
σ 2 ϵ 0 ( 4 π arctan 1 + a 2 2 z 2 − 1 ) {\displaystyle {\frac {\sigma }{2\epsilon _{0}}}({\frac {4}{\pi }}\arctan {\sqrt {1+{\frac {a^{2}}{2z^{2}}}}}-1)}
It is the same as:
σ π ϵ 0 arctan ( a 2 4 z z 2 + a 2 2 ) {\displaystyle {\frac {\sigma }{\pi \epsilon _{0}}}\arctan({\frac {a^{2}}{4z{\sqrt {z^{2}+{\frac {a^{2}}{2}}}}}})}
σ 2 ϵ 0 ( 4 π arctan 1 + a 2 2 z 2 − 1 ) = σ π ϵ 0 arctan ( a 2 4 z z 2 + a 2 2 ) {\displaystyle {\frac {\sigma }{2\epsilon _{0}}}({\frac {4}{\pi }}\arctan {\sqrt {1+{\frac {a^{2}}{2z^{2}}}}}-1)={\frac {\sigma }{\pi \epsilon _{0}}}\arctan({\frac {a^{2}}{4z{\sqrt {z^{2}+{\frac {a^{2}}{2}}}}}})}
1 2 ( 4 arctan 1 + a 2 2 z 2 − π ) = arctan ( a 2 4 z z 2 + a 2 2 ) {\displaystyle {\frac {1}{2}}(4\arctan {\sqrt {1+{\frac {a^{2}}{2z^{2}}}}}-\pi )=\arctan({\frac {a^{2}}{4z{\sqrt {z^{2}+{\frac {a^{2}}{2}}}}}})}
2 arctan 1 + a 2 2 z 2 − π 2 = arctan ( a 2 4 z z 2 + a 2 2 ) {\displaystyle 2\arctan {\sqrt {1+{\frac {a^{2}}{2z^{2}}}}}-{\frac {\pi }{2}}=\arctan({\frac {a^{2}}{4z{\sqrt {z^{2}+{\frac {a^{2}}{2}}}}}})}
Now take the Tangent of the left hand side:
tan ( 2 arctan 1 + a 2 2 z 2 − π 2 ) = − cot ( 2 arctan 1 + a 2 2 z 2 ) = − 1 − ( 1 + a 2 2 z 2 ) 2 1 + a 2 2 z 2 = a 2 4 z 2 1 + a 2 2 z 2 = a 2 4 z z 2 + a 2 2 {\displaystyle \tan(2\arctan {\sqrt {1+{\frac {a^{2}}{2z^{2}}}}}-{\frac {\pi }{2}})=-\cot(2\arctan {\sqrt {1+{\frac {a^{2}}{2z^{2}}}}})=-{\frac {1-(1+{\frac {a^{2}}{2z^{2}}})}{2{\sqrt {1+{\frac {a^{2}}{2z^{2}}}}}}}={\frac {a^{2}}{4z^{2}{\sqrt {1+{\frac {a^{2}}{2z^{2}}}}}}}={\frac {a^{2}}{4z{\sqrt {z^{2}+{\frac {a^{2}}{2}}}}}}}
This equals to the tangent of the right hand side.
