# did i get this right?

• August 21st 2013, 04:16 PM
mikeythemaniac
did i get this right?
my question is to write an expression that is quivalent to csc(2cot^-1(x)), that does not use trigonometric functions. I setup a triangle with the sides of 1, x, and the hypotenuse would be sqrt(1+x^2). after working through it i got: csc(2cot^1(x)) = (1 + x^2)/(2x)

if anyone can let me know if that's right i would really appreciate it, if not i can put my work down to try to see where i went wrong. thanks :)
• August 21st 2013, 08:15 PM
Soroban
Re: did i get this right?
Hello, mikeythemaniac!

Quote:

$\text{Write an expression that is equivalent to }\,\csc\left[2\cot^{\text{-}1}(x)\right].$

$\text{I set up a triangle with the sides of }1, x\,\text{ and the hypotenuse would be }\sqrt{1+x^2}$

$\text{After working through it, I got: }\csc\left[2\cot^{\text{-}1}(x)\right] \:=\: \frac{1 + x^2}{2x}$

Correct! . . . Good work!