Show that for $\mathit{k}\mathbf{=}\mathbf{0}$,$\mathit{u}\mathbf{=}\mathit{F}\mathbf{(}\mathit{\varphi }\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\mathit{\varphi }\mathbf{,}\mathit{s}\mathit{n}\mathit{u}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{u}\mathbf{,}\mathit{c}\mathit{n}\mathit{u}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{u}\mathbf{,}\mathit{d}\mathit{n}\mathit{u}\mathbf{=}\mathbf{1}\mathbf{.}$and for$\mathit{k}\mathbf{=}\mathbf{1}$$\mathit{u}\mathbf{=}\mathit{F}\mathbf{(}\mathit{\varphi }\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{=}\mathit{l}\mathit{n}\mathbf{(}\mathit{s}\mathit{e}\mathit{c}\mathit{\varphi }\mathbf{+}\mathit{t}\mathit{a}\mathit{n}\mathit{\varphi }\mathbf{)}\mathbf{,}\mathit{\varphi }\mathbf{=}\mathit{g}\mathit{d}\mathit{u}\mathbf{,}\mathit{s}\mathit{n}\mathit{u}\mathbf{=}\mathit{t}\mathit{a}\mathit{n}\mathit{h}\mathit{u}\mathbf{,}\mathit{c}\mathit{n}\mathit{u}\mathbf{=}\mathit{d}\mathit{n}\mathit{u}\mathbf{=}\mathbf{sech}\mathit{u}$.

The statements given are mentioned below:

$$\begin{gathered} u=F(\varphi ,k) \\ u=\underset{0}{\overset{\varphi }{\int }}\frac{1}{\sqrt{1-k^2{\sin }^2\theta }}d\theta \\ u=s{n}^{-1}\left(\sin \varphi \right) \end{gathered}$$

The elliptic form of the integral is defined as $\mathit{K}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\frac{\mathbf{\pi }}{\mathbf{2}}}{\mathbf{\int }}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }}}\mathit{d}\mathit{\theta }$.

For $k=0$.

$$\begin{gathered} u=F(\varphi ,0) \\ u=\underset{0}{\overset{\varphi }{\int }}d\theta \\ u=\varphi \end{gathered}$$

It is given that$u=s{n}^{-1}\left(\sin \varphi \right)$

Substitute $u=\varphi$in the above equation.

$$\begin{gathered} u=s{n}^{-1}\left(\sin \varphi \right) \\ snu=\sin \varphi \\ snu=\sin u \end{gathered}$$

Formula states that $cnu=\sqrt{1-{\left(\sin \varphi \right)}^2}$.

Simplify the formula mentioned above.

$$\begin{gathered} cnu=\sqrt{1-{\left(\sin \varphi \right)}^2} \\ cnu=\sqrt{1-{\left(\sin u\right)}^2} \\ cnu=\cos u \end{gathered}$$

Formula states that $dnu=\sqrt{1-{\left(k\sin \varphi \right)}^2}$

Simplify the formula mentioned above.

$$\begin{gathered} dnu=\sqrt{1-{\left(k\sin \varphi \right)}^2} \\ dnu=1 \end{gathered}$$

For$k=1$.

$$\begin{gathered} u=F(\varphi ,1) \\ u=\underset{0}{\overset{\varphi }{\int }}\frac{1}{\cos \theta }d\theta \\ u=\ln (\sec \varphi +\tan \varphi ) \end{gathered}$$

It is proved earlier that$snu=\sin \varphi$.

Formula states that $\tanh u=\sin \varphi$.

Hence, $snu=\tanh u$.

Formula states that $cnu=\sqrt{1-s{n}^{2}u}$.

$$\begin{gathered} cnu=\sqrt{1-s{n}^{2}u} \\ cnu=\sqrt{1-{\tanh }^{2}u} \\ cnu=\frac{1}{\cosh u} \\ cnu=sechu \end{gathered}$$

Formula states that $dnu=\sqrt{1-s{n}^{2}u}$.

$dnu=sechu$

Hence, the statements have been proven.