prove that: cos[tan^-1 {sin ( cot^-1 x)}] = {(1 + x^2)/(2 + x^2)}^1/2
Draw a right triangle with angle $\displaystyle \theta$, "near side" x, and "opposite side" 1. That way $\displaystyle cot(\theta)=$ "near side"/"opposite side"$\displaystyle = x/1= x$ so $\displaystyle sin(cot^{-1} x)= sin(\theta)$= "opposite side"/"hypotenuse". The "opposite side" is 1 and, by the Pythagorean theorem, the "hypotenuse" is $\displaystyle \sqrt{1+ x^2}$:
$\displaystyle sin(cot^{-1}(x))= \frac{1}{\sqrt{1+ x^2}}$.
Now, draw a right triangle with angle $\displaystyle \phi$, "opposite side" 1 and "near side" $\displaystyle \sqrt{1+ x^2}$ so that $\displaystyle tan(\phi)= \frac{1}{\sqrt{1+ x^2}}= sin(cot^{-1}(x))$ and so $\displaystyle \theta= tan^{-1}(sin(cot^{-1}(x)))$.
$\displaystyle cos(tan^{-1}(sin(cot^{-1}(x))))$ is the "near side"/"hypotenuse" for this triangle. Since the "near side" is $\displaystyle \sqrt{1+ x^2}$ and the "opposite side" is 1, use the Pythagorean theorem again to find the hypotenuse of this triangle.
If x = cot θ, adjacent side is x andopposite side is 1. So the hypotenuse is
$\displaystyle \sqrt(1+x^2)$ And
$\displaystyle \sin \theta = \frac{1}{\sqrt{1+x^2}}$
$\displaystyle \cot^{-1}{x} = \sin^{-1}\frac{1}{\sqrt{1+x^2}}$
$\displaystyle \sin(\cot^{-1}{x}) = \sin(\sin^{-1}\left( \frac{1}{\sqrt{1+x^2}} \right)$
= $\displaystyle \frac{1}{\sqrt{1+x^2}}$.
Similarly do the rest of the part.
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