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

Draw a right triangle with angle , "near side" x, and "opposite side" 1. That way "near side"/"opposite side" so = "opposite side"/"hypotenuse". The "opposite side" is 1 and, by the Pythagorean theorem, the "hypotenuse" is :
.

Now, draw a right triangle with angle , "opposite side" 1 and "near side" so that and so .

is the "near side"/"hypotenuse" for this triangle. Since the "near side" is and the "opposite side" is 1, use the Pythagorean theorem again to find the hypotenuse of this triangle.

Originally Posted by freetibet

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
And







= .

Similarly do the rest of the part.

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