TITLE: Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$
QUESTION [30 upvotes]: 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 $x$)?

REPLY [2 votes]: initially i  have
$\sin x+\cos x={7\over5}$
took the squares,
$\sin^2x+\cos^2x+2\sin x\cos x={49\over 25}$
$1+2\sin x\cos x={49\over 25}$
$\sin x\cos x={49\over 50}-{1\over 2}={12\over 25}$
$$\frac{1}{\sin x}+\frac{1}{\cos x}$$
$$\frac{\sin x+\cos x}{\sin x\cos x}$$
$$\frac{{7\over 5}}{{12\over 25}}$$
$${35\over 12}$$