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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 $x$)?

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10 Answers

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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 $\sin x+\cos x$, $$=2\cdot \frac{\frac 75}{\left(\frac{7}{5}\right)^2-1}$$ $$=\frac{35}{12}$$

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$$\sin x+\cos x=\frac{7}{5}$$ 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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\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}

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$$ s + c = \frac{7}{5};\quad s^2 + c^2 \frac{35}{12} = 1; $$

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:

$$ \frac{1}{s} + \frac{1}{c} =\frac{5}{4} +\frac{5}{3} = \frac{35}{12}. $$

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$$ \begin{align} \sin(x)+\cos(x)=\frac75 &\implies1-\sin^2(x)=\frac{49}{25}-\frac{14}5\sin(x)+\sin^2(x)\\ &\implies\sin^2(x)-\frac75\sin(x)+\frac{12}{25}=0\\ &\implies\sin(x)=\frac{7\pm1}{10}\\ &\implies\sin(x)\in\left\{\frac35,\frac45\right\} \end{align} $$ Since $\cos(x)=\frac75-\sin(x)\gt0$, if $\sin(x)=\frac35$ then $\cos(x)=\frac45$ and vice-versa. Therefore, $$ \frac1{\sin(x)}+\frac1{\cos(x)}=\frac{35}{12} $$

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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

$$b={\frac{\frac{14}{5}}{\frac{49}{25}-1}}={\frac{70}{24}}={\frac{35}{12}}$$

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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, $$\frac1{\sin x}+\frac1{\cos x}=\frac1a+\frac1b$$$$=\frac{1}{3/5}+\frac{1}{4/5}$$ $$=\frac{35}{12}$$

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initially i have

$\sin x+\cos x={7\over5}$

took the squares,

$\sin^2x+\cos^2x+2\sin x\cos x={49\over 25}$

$1+2\sin x\cos x={49\over 25}$

$\sin x\cos x={49\over 50}-{1\over 2}={12\over 25}$

$$\frac{1}{\sin x}+\frac{1}{\cos x}$$$$\frac{\sin x+\cos x}{\sin x\cos x}$$$$\frac{{7\over 5}}{{12\over 25}}$$$${35\over 12}$$

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$$ x+y= p\tag1$$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$$\;p= \dfrac{7}{5} $$

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$$\sin x+\cos x=\frac75\tag1$$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!

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