Let [tex]y=\tan^{-1}x[/tex], so that [tex]\tan y=x[/tex]. Picture a right triangle with hypotenuse of length [tex]z[/tex] and a reference angle [tex]y[/tex]. Then
[tex]\cos y=\dfrac 1z[/tex]
We have
[tex]\tan^2y=\sec^2y-1\implies x^2=z^2-1\implies z=\sqrt{x^2+1}[/tex]
[tex]\implies\cos(\tan^{-1}x)=\cos y=\dfrac1{\sqrt{x^2+1}}[/tex]
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Similar strategy: Let [tex]y=\arcsin x[/tex], so that [tex]\sin y=x[/tex]. In a right triangle with hypotenuse [tex]z[/tex] and reference angle [tex]y[/tex], we have
[tex]\cos y=\sqrt{1-x^2}[t/ex]
[tex]\implies\cot(\arcsin x)=\cot y=\dfrac{\cos y}{\sin y}=\dfrac{\sqrt{1-x^2}}x[/tex]
