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 $x$)?

$$ s + c = \frac{7}{5};\quad  s^2 + c^2 \frac{35}{12} = 1; $$
Solving quadratic equation by elimination of one of the two variables
$$ s= \left(\frac{4}{5}, \frac{3}{5} \right );\quad c = \left(\frac{3}{5}, \frac{4}{5} \right) ; $$
respectively. They are interchangeable. So only one result is obtained with either of two inputs:
$$ \frac{1}{s} + \frac{1}{c} =\frac{5}{4} +\frac{5}{3} = \frac{35}{12}. $$
