ランデン変換
i sn  ( u , k ) cn  ( u , k ) = 2 i 1 + k sn  ( 1 + k 2 u , 2 k 1 + k ) cn  ( 1 + k 2 u , 2 k 1 + k ) dn  ( 1 + k 2 u , 2 k 1 + k ) 2 1 + k dn 2  ( 1 + k 2 u , 2 k 1 + k ) − 1 − k 1 + k 4 k ( 1 + k ) 2 dn  ( 1 + k 2 u , 2 k 1 + k ) = 4 k i ( 1 + k ) 2 sn  ( 1 + k 2 u , 2 k 1 + k ) cn  ( 1 + k 2 u , 2 k 1 + k ) dn 2  ( 1 + k 2 u , 2 k 1 + k ) − 1 − k 1 + k {\displaystyle {\begin{aligned}{\frac {i\operatorname {sn} \left(u,k\right)}{\operatorname {cn} \left(u,k\right)}}&={\frac {\frac {{\tfrac {2i}{1+k}}\operatorname {sn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)\operatorname {cn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}{\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}{\frac {{\tfrac {2}{1+k}}\operatorname {dn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)-{\tfrac {1-k}{1+k}}}{{\tfrac {4k}{(1+k)^{2}}}\operatorname {dn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}}}\\&={\frac {{\tfrac {4ki}{(1+k)^{2}}}\operatorname {sn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)\operatorname {cn} \left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)}{\operatorname {dn} ^{2}\left({\tfrac {1+k}{2}}u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)-{\tfrac {1-k}{1+k}}}}\\\end{aligned}}}
sn  ( ( 1 + k ) u , 2 k 1 + k ) = ( 1 + k ) sn  α 1 + k sn 2  α {\displaystyle \operatorname {sn} \left((1+k)u,{\tfrac {2{\sqrt {k}}}{1+k}}\right)={\frac {(1+k)\operatorname {sn} \alpha }{1+k\operatorname {sn} ^{2}\alpha }}}
であるが、 u {\displaystyle u} を u 1 + k {\displaystyle {\tfrac {u}{1+k}}} に改め、 k {\displaystyle k} を 1 − 1 − k 2 1 + 1 − k 2 {\displaystyle {\tfrac {1-{\sqrt {1-k^{2}}}}{1+{\sqrt {1-k^{2}}}}}} に改めれば
F ( sin  α , k ) = 2 1 + k F ( 1 2 ( 1 + k ) 2 sin 2  α + ( 1 − k 2 sin 2  φ − 1 − sin 2  φ ) 2 , 2 k 1 + k ) {\displaystyle F\left(\sin \alpha ,k\right)={\frac {2}{1+k}}F\left({\frac {1}{2}}{\sqrt {\left(1+k\right)^{2}\sin ^{2}\alpha +\left({\sqrt {1-k^{2}\sin ^{2}\phi }}-{\sqrt {1-\sin ^{2}\phi }}\right)^{2}}},{\frac {2{\sqrt {k}}}{1+k}}\right)}
同じく、次の恒等式をガウス変換という。
F ( sin  α , k ) = 1 1 + k F ( ( 1 + k ) sin  α 1 + k sin 2  α , 2 k 1 + k ) {\displaystyle F\left(\sin \alpha ,k\right)={\frac {1}{1+k}}F\left({\frac {(1+k)\sin \alpha }{1+k\sin ^{2}\alpha }},{\frac {2{\sqrt {k}}}{1+k}}\right)}