# Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$

$$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$$

I got no idea how to find the solution to this. Can someone put me on the right track?

Thank you very much.

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This is a number! What exactly do you want to solve here? – Dennis Gulko Jul 18 '12 at 16:23
Using the word "solve" too broadly is a frequent mistake. "Evaluating" or "finding" makes more sense here. One solves equations; one solves problems; one evaluates expressions. – Michael Hardy Jul 18 '12 at 21:00

Divide both terms by two and use the fact $\sin(30) = \frac{1}{2}$ and $\cos(30) = \frac{\sqrt{3}}{2}$. Then you just need to use the formulas for $\sin(a+b)$ and $\sin(a-b)$ to find the solution.

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+1 Pretty nice and simple hint to a rather poorly posed question. – DonAntonio Jul 18 '12 at 16:27
Note that the use of \sin{something} looks better than sin{something}. The same goes for the other trig functions. Otherwise, nice hint! – Joe Jul 18 '12 at 16:28
Thank you! got me on the right track. Solved it :) – Grozav Alex Ioan Jul 18 '12 at 16:33

We have \begin{eqnarray*} E&=&\frac{\cos 10^\circ-\sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\\ &=&4\frac{(1/2)\cos 10^\circ-(\sqrt{3}/2)\sin 10^\circ}{2\sin 10^\circ \cos 10^\circ}\\ &=&4\frac{\cos 60^\circ\cos 10^\circ-\sin 60^\circ\sin 10^\circ}{\sin 20^\circ}\\ &=&4\frac{\cos(60^\circ+10^\circ)}{\sin 20^\circ}\\ &=&4\frac{\cos 70^\circ}{\sin 20^\circ}\\ &=&4\frac{\sin 20^\circ}{\sin 20^\circ}\\ &=&4. \end{eqnarray*}

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$$\text{Let us check the value of }\frac a{\sin\theta}+\frac b{\cos\theta}$$

$$\frac a{\sin\theta}+\frac b{\cos\theta}=\frac{a\cos\theta+b\sin\theta}{\cos\theta\sin\theta}$$

Putting $a=r\sin\alpha,b=r\cos\alpha$ where $r>0$

Squaring & adding we get $r^2=a^2+b^2\implies r=+\sqrt{a^2+b^2}$

$$\implies \frac a{\sin\theta}+\frac b{\cos\theta}=\frac{2\sqrt{a^2+b^2}(\sin\theta\cos\alpha+\cos\theta\sin\alpha)}{\sin2\theta}$$ $$=2\sqrt{a^2+b^2}\cdot\frac{\sin(\theta+\alpha)}{\sin2\theta}\text{ as }\sin2\theta=2\sin\theta\cos\theta$$

Now, the solution of $P\sin x= Q\sin A$ is general intractable unless $P=0$ or $Q=0$ or $P=\pm Q\ne0$

Here the coefficients of $\sin(\theta+\alpha),\sin2\theta$ can not be $0$

So, either $\sin2\theta=\sin(\theta+\alpha)$ or $\sin2\theta=-\sin(\theta+\alpha)$

$$\begin{array}{|c|c|c|} \hline \text{ Case } & \sin2\theta=\sin(\theta+\alpha) & \sin2\theta=-\sin(\theta+\alpha)=\sin(-\theta-\alpha) \text{ as }\sin(-x)=-\sin x \\ \hline \text{General Solution} & 2\theta=n180^\circ+(-1)^n(\theta+\alpha)\text{ where }n\text{ is any integer } & 2\theta=n180^\circ+(-1)^n(-\alpha-\theta)\text{ where }n\text{ is any integer } \\ \hline n=2m & \alpha=\theta-m360^\circ\equiv\theta\pmod{360^\circ} & \alpha=m360^\circ-3\theta\equiv-3\theta \\ \hline n=2m+1 & \alpha=(2m+1)180^\circ-3\theta\equiv 180^\circ-3\theta & \alpha=\theta-(2m+1)180^\circ\equiv\theta+180^\circ \\ \hline \end{array}$$

Here $a=1,b=-\sqrt3$ and $\theta=10^\circ$

Taking $\sin2\theta=\sin(\theta+\alpha), \alpha=\theta=10^\circ$ or $=180^\circ-3\theta=150^\circ$

$\implies r=+\sqrt{a^2+b^2}=2$ and $\cos \alpha=\frac br=-\frac{\sqrt3}2$ and $\sin\alpha=\frac ar=\frac12\implies \alpha$ lies in the 2nd Quadrant, $\implies \alpha=150^\circ$

$$\text{So,} \frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}=2\sqrt{1^2+(-\sqrt3)^2}\cdot\frac{\sin(10^\circ+150^\circ)}{\sin(2\cdot10^\circ)}=2\cdot2=4$$

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Could somebody please help me to show the left most column which is invisible in the answer – lab bhattacharjee Apr 23 '13 at 16:08