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*<math>\operatorname{Tanc}(z)= \frac {2i \operatorname{HeunB} \left( 2,0,0,0,\sqrt {2}\sqrt {iz} \right) }{(2z+\pi) \operatorname{HeunB} \left( 2,0,0,0,\sqrt {2}\sqrt {(i/2) (2z+\pi) } \right) } </math>
* <math>\operatorname{Tanc}(z)= \frac {{\rm WhittakerM}(0,\,1/2,\,2\,iz)}{{\rm WhittakerM}(0,\,1/2,\,i (2z+\pi)) z}
</math>
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: <math>\int _0^z \frac {\tan(x) }{x} \, dx = \left(z+ \frac {1}{9} z^3 + \frac {2}{75} z^5 + \frac {17}{2205} z^7 + \frac {62}{25515} z^9+ \frac {1382}{1715175} z^{11}+ \frac {21844}{
79053975} z^{13} + \frac{929569}{9577693125} z^{15}+ O (z^{17}) \right)</math>
==Pade approximation==
<math>{\it Tainc} \left( z \right) = \left( 1-{\frac {7}{51}}\,{z}^{2}+{
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^{8} \right) ^{-1}
</math>
==Gallery==
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