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 $x$)?

\begin{align}
  \sin(x)+\cos(x) &= \frac 75 \\
  \left(\sin(x)+\cos(x)\right)^2 &= \left( \frac 75 \right)^2 \\
  1 + 2 \sin(x) \cos(x) &= \frac{49}{25} \\
  \sin(x) \cos(x) &= \frac{12}{25}
\end{align}
\begin{align}
  \frac{1}{\sin(x)}  + \frac{1}{\cos(x)}
  &= \frac{\sin(x)  + \cos(x)}{\sin(x) \cos(x)}\\
  &= \frac{\left(\frac{7}{5}\right)}{\left(\frac{12}{25}\right)}\\
&= \frac{7 \times 25}{5\times 12}\\
  &= \frac{35}{12}
\end{align}
