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Calculate the exact values of the cot(theta) and sin(theta) for the acute angle theta if sec(theta)= the square root of 6.
Thank you so much!

sec Ø = √6
so cosØ = 1/√6
sketch a triangle with hypotenuse √6, adjacent = 1, let the opposite be y
1^2 + y^2 = √6^2
y^2 = 5
y = √5

then sinØ = √5/√6
cotØ = 1/√5

Could you tell me why you squared those numbers? Is there a formula?

If I was doing sin instead, would 1^2 +y^2=√6^2 be in a different order?

Also how did you get the 5? Isn't the square of 6 to the second power just 6?

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