User:Lightest

In the book:

$$ \frac{\sigma}{2\epsilon_0} ( \frac{4}{\pi} \arctan \sqrt{ 1+ \frac{a^2}{2z^2} } - 1 ) $$

It is the same as:

$$ \frac{\sigma}{\pi \epsilon_0} \arctan ( \frac{a^2}{4z\sqrt{z^2+\frac{a^2}{2}}} ) $$

$$ \frac{\sigma}{2\epsilon_0} ( \frac{4}{\pi} \arctan \sqrt{ 1+ \frac{a^2}{2z^2} } - 1 ) = \frac{\sigma}{\pi \epsilon_0} \arctan ( \frac{a^2}{4z\sqrt{z^2+\frac{a^2}{2}}} ) $$

$$ \frac{1}{2} ( 4 \arctan \sqrt{ 1+ \frac{a^2}{2z^2} } - \pi ) =  \arctan ( \frac{a^2}{4z\sqrt{z^2+\frac{a^2}{2}}} ) $$

$$ 2 \arctan \sqrt{ 1+ \frac{a^2}{2z^2} } - \frac{\pi}{2} =  \arctan ( \frac{a^2}{4z\sqrt{z^2+\frac{a^2}{2}}} ) $$

Now take the Tangent of the left hand side:

$$ \tan(2 \arctan \sqrt{ 1+ \frac{a^2}{2z^2} } - \frac{\pi}{2}) = -\cot(2 \arctan \sqrt{ 1+ \frac{a^2}{2z^2} }) = -\frac{1-(1+ \frac{a^2}{2z^2})}{2 \sqrt{1+ \frac{a^2}{2z^2}}} = \frac{a^2}{4z^2 \sqrt{1+ \frac{a^2}{2z^2}}} = \frac{a^2}{4z \sqrt{z^2+ \frac{a^2}{2}}}$$

This equals to the tangent of the right hand side.