Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
  \sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of 
\begin{equation}
  \frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) without using the identities of trigonometry.
The function $\sin x+\cos x$ could be transformed using some trigonometric identities to a single function. In fact, WolframAlpha says it is equal to $\sqrt2\sin\left(x+\frac\pi4\right)$ and there also are some posts on this site about this equality. So probably in this way we could calculate $x$ from the first equation - and once we know $\sin x$ and $\cos x$, we can calculate $\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$. Is there a simpler solution (perhaps avoiding explicitly finding $x$)?

$$ x+y= p\tag1$$
Square, since $( x^2+y^2=1 )$
$$ 1+ 2 x\;y = p^2, \; x y= \dfrac{p^2-1}{2} \tag2$$
From (1) and (2)
$$ \dfrac{1}{x}+  \dfrac{1}{y} = \dfrac{x+y}{x y}=  \dfrac{2p}{p^2-1} $$
$$ = \dfrac{35}{12},\;$$
if
$$\;p= \dfrac{7}{5} $$
