What is the exact value of $\frac{1}{\tan20^{\circ}}-\frac{1}{\sin80^{\circ}}$?
By inputting the above into a calculator, I get $\sqrt{3}$, but I cannot seem to be able to do it algebraically.
What is the exact value of $\frac{1}{\tan20^{\circ}}-\frac{1}{\sin80^{\circ}}$?
0
$\begingroup$
trigonometry
-
0https://www.wolframalpha.com/input/?i=1%2Ftan(20)+-1%2Ftan(80). Where do you see $\sqrt{3}$? – 2017-02-25
-
0https://www.wolframalpha.com/input/?i=tan(70+deg)-+tan(10+deg) $\approx 2.571$ – 2017-02-25
-
0Are you sure? Because WolframAlpha says that this is 2.57115... which is not $\sqrt{3}$ – 2017-02-25
-
0@mannav very sorry for this, it should be sin rather than tan in one of the fractions – 2017-02-25
1 Answers
0
In any case for computing the result you can use the following trigonometric identities:
$\tan(80^{\circ})=\tan(\frac{\pi}{3}+20^{\circ})=\frac{\tan(\frac{\pi}{3})+\tan(20^{\circ})}{1-\tan(\frac{\pi}{3})\tan(20^{\circ})}=\frac{\sqrt{3}+\tan(20^{\circ})}{1-\sqrt{3}\tan(20^{\circ})}$
Now, let's target $\tan(20^{\circ})$ to find also the value for $\tan(80^{\circ})$. We can write:
$\tan(60^{\circ})=\tan(20^{\circ}+20^{\circ}+20^{\circ})=\frac{3\tan(20^{\circ})-\tan^3(20^{\circ})}{1-3\tan^2(20^{\circ})}$
Setting $x=\tan(20^{\circ})$ in the last equation we obtain: $\sqrt{3}=\frac{3x-x^3}{1-3x^2}$. And hence (assuming $1-3x\ne 0$):
$x^3-3\sqrt{3}x^2-3x+\sqrt{3}=0$
I think you can go from here.
-
0Just changed the question so that it does give $\sqrt{3}$ – 2017-02-25