Question

cot(cos^-1u)
please help I just don't understand these questions!!

Answer 1

use the identity :
\(\large 1 + \tan^2x = \sec^2x\)

put \(x = \cos^{-1}u\)

Answer 2

\(\large 1 + \tan^2(\cos^{-1}u) = \frac{1}{\cos^2(\cos^{-1}u)}\)

\(\large 1 + \tan^2(\cos^{-1}u) = \frac{1}{u^2}\)

\(\large \tan^2(\cos^{-1}u) = \frac{1-u^2}{u^2}\)

\(\large \cot (\cos^{-1}u) = \frac{u}{\sqrt{1-u^2}}\)

Answer 3

Thank you so much for your help!! Trig is one confusing subject. Can you tell me what you did with tan^2 in the second to last part? No really understanding that part.

Answer 4

very good question :)

u knw tan = 1/cot right ?

Answer 5

\(\large \tan^2(\cos^{-1}u) = \frac{1-u^2}{u^2}\)

\(\large \frac{1}{\tan^2(\cos^{-1}u)} = \frac{u^2}{1-u^2}\)

\(\large \cot^2(\cos^{-1}u) = \frac{u^2}{1-u^2}\)

Answer 6

take square-root both sides

Answer 7

oh thank you for explaining. It makes so much more sense now. I should never take a online math class again.

Answer 8

glad it made some sense :)