\right) \left(
u\right) =\dfrac{u}{K\left( u,1\right) }\text{ }_{1}\Phi _{0}\left[ 
\begin{array}{c}
q \\ 
\end{array}\left\vert q,-\dfrac{au}{q}\right. \right] .\end{array}\right.$$Also, readers may easily verify that$$_{1}\Phi _{0}\left[ 
\begin{array}{c}
q \\ 
\end{array}\left\vert q,-\frac{au}{qv}\right. \right] =\sum\limits_{0}^{\infty }\left( 
\frac{au}{qv}\right) ^{n}=\dfrac{au}{qv+au},\text{ }\left\vert au\right\vert
<qv.  \tag{42}$$From above and the fact that $K\left( s,1\right) =1$ we have the following corrollary.

<span style="font-variant:small-caps;">Corrollary 13.</span> *Let* $a$* be a real number. Then, we have*$$_{q}N\left( E_{q}\left( ax\right) \right) \left( u,v\right) =\dfrac{qu}{\left( qv+au\right) },\text{ }\left\vert au\right\vert <qv.  \tag{43}$$

Hence, $\left( 43\right) $ indeed reveals

$\left( \mathtt{i}\right) $ $_{q}L\left( E_{q}\left( ax\right) \right)
\left( v\right) =\dfrac{q}{\left( qv+a\right) },$ $\left\vert a\right\vert
<qv;$ $\ \left( \mathtt{ii}\right) $ $_{q}S\left( E_{q}\left( ax\right)
\right) \left( u\right) =\dfrac{qu}{\left( q+au\right) },$ $\left\vert
au\right\vert <q.$

It may also be mentioned here that Corrollary 13 and the identities$$_{q}\sin x=\frac{E_{q}\left( ix\right) -E_{q}\left( -ix\right) }{2i}\text{
and }_{q}\cos x=\frac{E_{q}\left( ix\right) -E_{q}\left( -ix\right) }{2} 
\tag{44}$$state, without proof, the following result.

<span style="font-variant:small-caps;">Corrollary 14.</span> *Let* $a$* be a real number. Then, we have*

$\left( \mathtt{i}\right) $ $_{q}N\left( _{q}\sin ax\right) \left(
u;v\right) =\dfrac{-qau^{2}}{q^{2}v^{2}+a^{2}u^{2}},$ $\left\vert
au\right\vert <qv.$

$\left( \mathtt{ii}\right) $ $_{q}N\left( _{q}\cos ax\right) \left(
u;v\right) =\dfrac{q^{2}uv}{q^{2}v^{2}+a^{2}u^{2}},$ $\left\vert
au\right\vert <qv.$

Further, Corrollary 14 suggests to have the following conclusions proclaimed.

$\left( \mathtt{i}\right) $ $_{q}L\left( _{q}\sin ax\right) \left( v\right) =\dfrac{-qa}{q^{2}v^{2}+a^{2}},$ $\left\vert a\right\vert <qv;$ $\ \ \ 