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 $x$)?
 A: $$
\begin{align}
\sin(x)+\cos(x)=\frac75
&\implies1-\sin^2(x)=\frac{49}{25}-\frac{14}5\sin(x)+\sin^2(x)\\
&\implies\sin^2(x)-\frac75\sin(x)+\frac{12}{25}=0\\
&\implies\sin(x)=\frac{7\pm1}{10}\\
&\implies\sin(x)\in\left\{\frac35,\frac45\right\}
\end{align}
$$
Since $\cos(x)=\frac75-\sin(x)\gt0$, if $\sin(x)=\frac35$ then $\cos(x)=\frac45$ and vice-versa. Therefore,
$$
\frac1{\sin(x)}+\frac1{\cos(x)}=\frac{35}{12}
$$
A: $$ s + c = \frac{7}{5};\quad  s^2 + c^2 \frac{35}{12} = 1; $$
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:
$$ \frac{1}{s} + \frac{1}{c} =\frac{5}{4} +\frac{5}{3} = \frac{35}{12}. $$
A: If 
$$\sin x+\cos x=a\qquad\text{and}\qquad{\frac{1}{\sin x}}+{\frac{1}{\cos x}}=b$$ then 
$$a=b\sin x\cos x\qquad\text{and}\qquad a^2=1+2\sin x\cos x=1+\frac{2a}{b}$$
from which it follows that
$$b={\frac{2a}{a^2-1}}$$
Letting $a=\frac{7}{5}$, we have
$$b={\frac{\frac{14}{5}}{\frac{49}{25}-1}}={\frac{70}{24}}={\frac{35}{12}}$$
A: Assume that, 
$\sin x=a, \cos x=b $
Given that : $$\sin x+\cos x=\frac75$$ 
$$a+b=\frac75\tag 1$$ 
$$\sin^2 x+\cos^2 x=1$$ 
$$a^2 +b^2 =1$$ 
$$(a+b)^2-2ab =1$$ 
$$(7/5)^2-2ab =1$$ 
$$ab=12/25\tag 2$$ 
solving (1), (2), i get $a=3/5, b=4/5$
therefore,
$$\frac1{\sin x}+\frac1{\cos x}=\frac1a+\frac1b$$$$=\frac{1}{3/5}+\frac{1}{4/5}$$ $$=\frac{35}{12}$$
A: initially i  have
$\sin x+\cos x={7\over5}$
took the squares,
$\sin^2x+\cos^2x+2\sin x\cos x={49\over 25}$
$1+2\sin x\cos x={49\over 25}$
$\sin x\cos x={49\over 50}-{1\over 2}={12\over 25}$
$$\frac{1}{\sin x}+\frac{1}{\cos x}$$
$$\frac{\sin x+\cos x}{\sin x\cos x}$$
$$\frac{{7\over 5}}{{12\over 25}}$$
$${35\over 12}$$
A: $$\sin x+\cos x=\frac{7}{5}$$
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
$$\frac{5}{4}+\frac{5}{3}=\frac{35}{12}$$
A: \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}
A: Notice, $$\frac{1}{\sin x}+\frac{1}{\cos x}$$
$$=\frac{\sin x+\cos x}{\sin x\cos x}$$
$$=2\cdot \frac{\sin x+\cos x}{2\sin x\cos x}$$
$$=2\cdot \frac{\sin x+\cos x}{(\sin x+\cos x)^2-1}$$
setting the value of $\sin x+\cos x$, 
$$=2\cdot \frac{\frac 75}{\left(\frac{7}{5}\right)^2-1}$$
$$=\frac{35}{12}$$
A: $$\sin x+\cos x=\frac75\tag1$$
take the squares,
$$(\sin x+\cos x)^2=\left(\dfrac75\right)^2$$
$$\sin x\cos x=\frac{12}{25}\tag2$$
$$\therefore \dfrac{1}{\sin x}+\dfrac{1}{\cos x}$$
$$=\dfrac{\sin x+\cos x}{\sin x\cos x}$$
$$=\dfrac{\left(\dfrac{7}{5}\right)}{\left(\dfrac{12}{25}\right)}$$
$$=\dfrac{35}{12}$$
Hope it helps!
A: $$ x+y= p\tag1$$
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
$$\;p= \dfrac{7}{5} $$
