<s>[INST] 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 $x$)?
 [/INST] $$\sin x+\cos x=\frac{7}{5}$$
Let  $\sin x=t$
$$\implies t+\sqrt{1-t^2}=\frac{7}{5}$$
Shifting, squaring and simplifying, we get
$$25t^2-35t+12=0$$
$$\implies t=\frac{35 \pm 5}{50}$$
Hence, $$\sin x= \frac{4}{5},\ \cos x=\frac{3}{5} \ \text{or} \ \cos x= \frac{4}{5}, \ \sin x=\frac{3}{5}$$
But as we need to find $$\frac{1}{\sin(x)}  +\frac{1}{\cos(x)}$$ it becomes
$$\frac{5}{4}+\frac{5}{3}=\frac{35}{12}$$
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