$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$
I got no idea how to find the solution to this. Can someone put me on the right track?
Thank you very much.
$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$
I got no idea how to find the solution to this. Can someone put me on the right track?
Thank you very much.
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.
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*}
$\text{Let us check the value of }\frac a{\sin\theta}+\frac b{\cos\theta}$
$\frac a{\sin\theta}+\frac b{\cos\theta}=\frac{a\cos\theta+b\sin\theta}{\cos\theta\sin\theta}$
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, $\implies \alpha=150^\circ$
$\text{So,} \frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}=2\sqrt{1^2+(-\sqrt3)^2}\cdot\frac{\sin(10^\circ+150^\circ)}{\sin(2\cdot10^\circ)}=2\cdot2=4$