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May 2010
5
1
prove that: cos[tan^-1 {sin ( cot^-1 x)}] = {(1 + x^2)/(2 + x^2)}^1/2
 

HallsofIvy

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Apr 2005
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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.
 
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Jun 2009
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prove that: cos[tan^-1 {sin ( cot^-1 x)}] = {(1 + x^2)/(2 + x^2)}^1/2
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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May 2010
5
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dear Mr. sa ri ga ma:
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