Question

1)is this rite ? CAN YOU SAY HOW ?
=(1-cosA/1+cosA) ^1/2
=1-COSA/SINA

Answer 1

I can see it... you ready? ^^

Answer 2

ya

Answer 3

Okay, pay attention.
\[\Large \left[\frac{1-\cos A}{1+ \cos A}\right]^{\frac12}\]

Answer 4

oaky

Answer 5

We'll multiply the inside of the bracket by 1, which is ok, except we multiply a rather special variant of "1".
Namely
\[\large \frac{1-\cos(A)}{1-\cos(A)}\]

Answer 6

So let's do it.
\[\Large \left[\frac{1-\cos A}{1+ \cos A}\cdot \frac{1-\cos A}{1-\cos A}\right]^{\frac12}\]
Do you see where this is going?

Answer 7

okay

Answer 8

Can you take it from here?

Answer 9

yes i got

Answer 10

So, we're done? ^_^

Answer 11

but after i want to get
=1-cosA/sinA

Answer 12

Okay, relax.
Simplify the multiplication, tell me what you get.
\[\Large \left[\frac{(1-\cos A)(1-\cos A)}{(1+ \cos A)(1-\cos A)}\right]^{\frac12}\]

Answer 13

this is my question

Answer 14

I know, but I'm asking you to simplify the multiplication above. ^

You can do it :)

Answer 15

yes i can do.. after i got finely
=[(1−cosA)(1−cosA)(1+cosA)(1−cosA)]12
=[1−cosA1+cosA]12

Answer 16

No, lol... write it down. I can't see clearly when it's all typed out like that

Answer 17

=[(1−cosA)(1−cosA)(1+cosA)(1−cosA)]12
=[1−cosA1+cosA]12
=1-COSA/sinnA
=(1/SINA)-(COSA/SINA)
=cosecA*cotA
=rhs

is this okay my friend?

Answer 18

No, it's like this... one step at a time
\[\Large \left[\frac{(1-\cos A)^2}{1- \cos^2 A}\right]^{\frac12}\]

Can you now see what's going to happen?

Answer 19

oh i got this.. [(1-cosa)^2/1-cos^a]^1/2

Answer 20

You forgot the ^2 in the cos in the denominator.

Anyway, we can't do anything about the numerator now, but the denominator... you know from Pythagorean identities that that must be equal to...?

Answer 21

yes now i can write this.. (1-cosA/1+cos^0A )^1/2

Answer 22

Oh... no, you can't.
LOL

Answer 23

\[\Large 1-\cos^2A = \color{red}?\]
Pythagorean identies... look back.

Answer 24

sin^2A

Answer 25

That is correct. So denominator now becomes...
\[\Large \left[\frac{(1-\cos A)^2}{\sin^2A}\right]^{\frac12}\]

Now just get the square root of the whole thing.

Finish it!
LOL

Answer 26

terenzreignz
oh my god.. thanks you it so ezy now.........lol

Answer 27

No problem :)
And call me TJ next time, ok?
Signing off now
-----------------------------
Terence out