# [SOLVED] person who helped will be highly appreciated

#### freetibet

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

#### HallsofIvy

MHF Helper
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.

• freetibet

#### sa-ri-ga-ma

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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• freetibet

#### freetibet

dear Mr. sa ri ga ma:
your reply(answer) to my first question on this site have fully satisfy me. i really appreciate your cooperation and help. thank you very much.

i really appreciate your valuable ideas and anwer, your's answer fully satisfied me..thank you

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