Show that cotθ = (sqrt(1 - x^2))/x please help

[alex1234]

I understand the question, however I don't quite know how to prove it.
Question:
Given that θ = arcsin x and that 0 < θ < pie/2 show that
cotθ = (sqrt(1 - x^2))/x, 0 < x < 1

Edit: I have figured it out myself now never mind, however for anyone interested it was actually pretty easy,

Given θ = arcsin x,
x = sin θ,
cot θ ≡ cos θ / sin θ
cos θ ≡ sqrt(1 - sin^2 θ)
So, cot θ = sqrt(1 - sin^2 θ) / sin θ
We said that x = sin θ, So
cot θ = sqrt(1 - x^2) / x

 

[stapel]

Given that θ = arcsin x and that 0 < θ < pie/2 show that
cotθ = (sqrt(1 - x^2))/x, 0 < x < 1

Edit: I have figured it out myself now never mind, however for anyone interested it was actually pretty easy,

Given θ = arcsin x,
x = sin θ,
cot θ ≡ cos θ / sin θ
cos θ ≡ sqrt(1 - sin^2 θ)
So, cot θ = sqrt(1 - sin^2 θ) / sin θ
We said that x = sin θ, So
cot θ = sqrt(1 - x^2) / x

Thank you for returning and posting your solution!

 

[HallsofIvy]

An "informal" proof: since \(\displaystyle sin(\theta)\) is "opposite side over hypotenuse", draw a right triangle with one angle labeled "\(\displaystyle \theta\), the opposite side "x", and the hypotenuse "1". By the Pythagorean theorem, the "near side" has length \(\displaystyle \sqrt{1- x^2}\). Then \(\displaystyle cot(\theta)\), "near side divided by opposite side", is \(\displaystyle \frac{\sqrt{1- x^2}}{x}\).

 

[alex1234]

An "informal" proof: since \(\displaystyle sin(\theta)\) is "opposite side over hypotenuse", draw a right triangle with one angle labeled "\(\displaystyle \theta\), the opposite side "x", and the hypotenuse "1". By the Pythagorean theorem, the "near side" has length \(\displaystyle \sqrt{1- x^2}\). Then \(\displaystyle cot(\theta)\), "near side divided by opposite side", is \(\displaystyle \frac{\sqrt{1- x^2}}{x}\).

Yeah my teacher ended up handing the mark scheme out both ways were valid, and honestly your way is better for less mathematical people as it requires less knowledge of trigonometric identities, however my way looks fancier