Question

(cot^2u- 1+ csc^2 u) / secu

Answer 1

What is the question? What is the goal?

Answer 2

To simplify but using fundemental identities

Answer 3

By*

Answer 4

Okay, well then I need to be sure of what it is. \[ \Large
\frac{\cot^2(u) - 1 + \csc^2(u)}{\sec(u)}
\]

Answer 5

Is this correct?

Answer 6

Yes

Answer 7

First identity to remember is: \[ \large
\cot^2(u) + 1 = \csc^2(u)
\]

Answer 8

Okay well i came jp with cotu+cscu+csc^2u but idk if thats right

Answer 9

This is the pythagorean theorem... comes from \[
\sin^2(u) + \cos^2(u) = 1
\] and dividing the whole thing by \(\sin^2(u)\).

Answer 10

So is my answer right?

Answer 11

You want to use that identity to get rid of the \(csc^2(u)\). Can you do that?

Answer 12

Well csc^2 is also 1/sin^2

Answer 13

No, what I mean is, replace the csc^2 with cot^2 + 1

Answer 14

Ohhh ok i see where i messsed up

Answer 15

Did you get the answer?

Answer 16

I got (sec)(secu+sin^2+1)???

Answer 17

Wait no..

Answer 18

(Cosu)(cosu+cos^2u+1)

Answer 19

\[
\Large

\begin{array}{l}
\frac{\cot^2(u) - 1 + \csc^2(u)}{\sec(u)} \\\\
\frac{\cot^2(u)-1+\cot^2(u)+1}{\sec(u)} \\\\
\frac{2\cot^2(u)}{\sec(u)} \\\\
\frac{2\cot^2(u)\cos(u)}{\sec(u)\cos(u)} \\\\
2\cot^2(u)\cos(u)\\\\
\frac{2\cos^3(u)}{\sin^2(u)}

\end{array}
\]

Answer 20

Thank you !!!!!!