[tex]\bf cot[cos^{-1}(u)]\\\\
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\textit{well, if }cos^{-1}(u)\textit{ give us an angle of say }\theta
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\textit{that simply means that }cos(\theta)=u\iff cos(\theta)=\cfrac{u}{1}\\\\
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cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad cos(\theta)=\cfrac{u}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}
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[tex]\bf \textit{let's find the value of "b", or the opposite side}
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\textit{using the pythagorean theorem}
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c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\implies \sqrt{1^2-u^2}=b\\\\\\ \sqrt{1-u^2}=b\\\\
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thus \qquad cot[cos^{-1}(u)]\implies cot(\theta)\implies cot(\theta)=\cfrac{adjacent}{opposite}
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cot(\theta)=\cfrac{u}{\sqrt{1-u^2}}[/tex]
