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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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    $\begingroup$ Note: $ \sin x+ \cos x = \sqrt2\sin(x+\frac{\pi}{4})$ $\endgroup$
    – George
    Commented Feb 10, 2016 at 20:30
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    $\begingroup$ It is impossible without identities. $\endgroup$
    – N.S.JOHN
    Commented Feb 26, 2016 at 15:37
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    $\begingroup$ my hint: $(\sin x+\cos x)^2=1+2\sin x\cosx$ $\endgroup$ Commented Feb 27, 2016 at 12:13
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    $\begingroup$ If this is the sort of post that ends up becoming a HNQ, I think it really, really demeans this site. The entire SE network can just click on the HNQ, and assume this question, and it's answers which are easily found via Wolfram Alpha, represent this sites quality. In this case, not at all. $\endgroup$
    – amWhy
    Commented Aug 6, 2017 at 16:47

10 Answers 10

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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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    $\begingroup$ thanks mate , that was pretty easy $\endgroup$ Commented Feb 10, 2016 at 20:33
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    $\begingroup$ You used the trigonometric identity $sin^(x)+cos^2(x)=1$ $\endgroup$ Commented Oct 23, 2019 at 10:57
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    $\begingroup$ 100 up votes :) $\endgroup$
    – Wolgwang
    Commented Dec 19, 2021 at 17:46
  • $\begingroup$ Just wondering ... Suppose we let $\sin(x)+\cos(x)=a$ then we'll get $$\frac{1}{\cos(x)}+\frac{1}{\sin(x)}=2\cdot\frac{a}{a^2-1},$$ even if $a=10$, say? $\endgroup$ Commented Dec 27, 2022 at 10:55
  • $\begingroup$ No. It is valid only for $-\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$ . Therefore the range of $a$ must be $-\sqrt{2}\le a\le \sqrt{2}$. Hence the formula: $\dfrac{1}{\cos x}+\dfrac{1}{\sin x}=2\cdot \dfrac{a}{a^2-1}$ is not valid for $a=10$ $\endgroup$ Commented Dec 27, 2022 at 11:20
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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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$$ \begin{align} \sin(x)+\cos(x)=\frac75 &\implies1-\sin^2(x)=\frac{49}{25}-\frac{14}5\sin(x)+\sin^2(x)\\ &\implies\sin^2(x)-\frac75\sin(x)+\frac{12}{25}=0\\ &\implies\sin(x)=\frac{7\pm1}{10}\\ &\implies\sin(x)\in\left\{\frac35,\frac45\right\} \end{align} $$ 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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$$ s + c = \frac{7}{5};\quad s^2 + c^2 \frac{35}{12} = 1; $$

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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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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    $\begingroup$ Approved your nice solution. +1. $\endgroup$
    – Sebastiano
    Commented Jun 28, 2020 at 9:46
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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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    $\begingroup$ Your answer adds nothing new to the already existing answers. $\endgroup$ Commented May 2, 2020 at 19:10
  • $\begingroup$ I answered it in my own way. $\endgroup$
    – user766881
    Commented May 2, 2020 at 23:34

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