[tex]cot(\frac{-11\pi}{6} )=\sqrt{3}[/tex]
What is reference angle?
"An acute angle enclosed between the terminal arm and the x-axis."
For given question,
We need to find the value of [tex]cot(\frac{-11\pi}{6} )[/tex] using the reference angle.
We know that, [tex]cot\theta=\frac{cos\theta}{sin\theta}[/tex]
So, [tex]cot(\frac{-11\pi}{6})=\frac{cos(\frac{-11\pi}{6})}{sin(\frac{-11\pi}{6})}[/tex] .................(1)
First we find the value of [tex]sin(\frac{-11\pi}{6})[/tex]
We know that [tex]sin(-\theta)=-sin(\theta)[/tex]
⇒ [tex]sin(\frac{-11\pi}{6})=-sin(\frac{11\pi}{6})[/tex] ...............(2)
Now,
[tex]sin(\frac{11\pi}{6})\\\\=sin(2\pi-\frac{\pi}{6})\\\\=-sin(\frac{\pi}{6})\\\\=-\frac{1}{2}[/tex]
From (2),
[tex]sin(\frac{-11\pi}{6})\\\\=-(-\frac{1}{2} )\\\\=\frac{1}{2}[/tex] .............(3)
Similarly we find the value of [tex]cos(\frac{-11\pi}{6})[/tex]
We know that [tex]cos(-\theta)=cos(\theta)[/tex]
[tex]\Rightarrow cos(\frac{-11\pi}{6})=cos(\frac{11\pi}{6})[/tex] ..........(4)
Now,
[tex]cos(\frac{11\pi}{6})\\\\=cos(2\pi-\frac{\pi}{6})\\\\=cos(\frac{\pi}{6})\\\\=\frac{\sqrt{3} }{2}[/tex]
From (4),
[tex]\Rightarrow cos(\frac{-11\pi}{6})\\\\=\frac{\sqrt{3} }{2}[/tex] ......................(5) .
From (1), (3) and (5),
[tex]cot(\frac{-11\pi}{6} )\\\\=\frac{\frac{\sqrt{3} }{2} }{\frac{1}{2} } \\\\=\sqrt{3}[/tex]
Therefore, [tex]cot(\frac{-11\pi}{6} )=\sqrt{3}[/tex]
Learn more about the reference angle here:
https://brainly.com/question/1603873
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