1. ## Proving confuction Identity

Hello,

Problem:

csc(pi/2 - u) = sec u

I understand that csc = 1/sin

how do I factor ----- 1/sin(pi/2-u)

That's where I'm basically stuck.

Let me know, thanks!

2. Originally Posted by l flipboi l
Hello,

Problem:

csc(pi/2 - u) = sec u

I understand that csc = 1/sin

how do I factor ----- 1/sin(pi/2-u)

That's where I'm basically stuck.

Let me know, thanks!
$\displaystyle csc \left(\frac{\pi}{2} - u \right) = \frac{1}{sin\left(\frac{\pi}{2} - u \right)}$

$\displaystyle sin(A+B) = sinAcosB + cosAsinB$

For $\displaystyle \theta = \frac{\pi}{2}$

$\displaystyle sin \theta = 1$ and $\displaystyle cos \theta = 0$

3. Originally Posted by l flipboi l
Hello,

Problem:

csc(pi/2 - u) = sec u

I understand that csc = 1/sin

how do I factor ----- 1/sin(pi/2-u)

That's where I'm basically stuck.

Let me know, thanks!
$\displaystyle \sin\left(\frac{\pi}{2} - u\right) = \sin\left(\frac{\pi}{2}\right)\cos(u) - \cos\left(\frac{\pi}{2}\right)\sin(u)$

continue ...

4. So the law of distributing fractions is:

a/b (c+d) = a / (bc+bd) ?

5. Originally Posted by l flipboi l
So the law of distributing fractions is:

a/b (c+d) = a / (bc+bd) ?
don't complicate things ...

just evaluate/simplify the expression

$\displaystyle \sin\left(\frac{\pi}{2} - u\right) = \sin\left(\frac{\pi}{2}\right)\cos(u) - \cos\left(\frac{\pi}{2}\right)\sin(u)$

and finish the proof.