Question

Prove the identity:
csc^2 x - 1 / cos x = cot x csc x
HELP ME PLEAASEEE. :D

Answer 1

What have you got to begin with? Want to start with the Pythagorean?

Answer 2

okay po. ahm, how po?

Answer 3

With this, to be precise...
cos² x + sin² x = 1

Now, if you divide both sides of the equation by
sin² x, you get
cot² x + 1 = csc² x

Answer 4

Please give me some indication that you are able to follow.

Answer 5

cot² x + 1 = csc² x uuse this formula

Answer 6

what po? i didn't get it po. sorry, im so slow po.

Answer 7

You know the Pythagorean identity
cos² x + sin² x = 1
?

Answer 8

@terenzreignz u can't use rhs side since here u hav to prove .did u gt wat i m saying

Answer 9

yes po .. i know that po ..

Answer 10

@manishsatywali I'm not using the rhs, I'm deriving the formula you just gave.

Answer 11

k pardon me............

Answer 12

Okay, well, starting from
cos² x + sin² x = 1
Well, divide both sides by sin² x, you get
\[\frac{\cos^{2}x}{\sin^{2}x}+\frac{\sin^{2}x}{\sin^{2}x}=\frac{1}{\sin^{2}x}\]
Right?

Answer 13

yes po. then po?

Answer 14

Okay, we can bring out the exponents, right?
\[\left( \frac{\cos x}{\sin x} \right)^{2}+1=\left( \frac{1}{\sin x}\right)^{2}\]
Okay?

Answer 15

okay po . then after that po?

Answer 16

What's cos x / sin x
?

Answer 17

cot x po .

Answer 18

Very good.
Now what's 1/sin x
?

Answer 19

csc x po .

Answer 20

So we can rewrite that equation as
cot² x + 1 = csc² x
am I right?

Answer 21

is it on the right side?

Answer 22

What's on the right side?
I just replaced cos x / sin x
with cot x
and 1/sin x
with csc x

Answer 23

ahh .. okays po . i get it . then po?

Answer 24

Oh, we're working on the left side, for now...

Answer 25

okay po ..

Answer 26

Okay, well, let me rewrite it
cot² x + 1 = csc² x
Then we can subtract 1 from both sides, and get
cot² x = csc² x - 1
Are you catching me so far?

Answer 27

yes i am ..

Answer 28

Well, then, you have
(csc² x - 1)/cos x
On the left side, right?
Slowly, we'll turn it into the right side...
First off, you know for a fact that
csc² x - 1
is equal to...?

Answer 29

cot^2 x po .

Answer 30

So we can turn the left side into
(cot² x) / cos x
right?

Answer 31

yes po.

Answer 32

Let me just write it more neatly:
\[\frac{\cot^2 x}{\cos x}=(\cot^2 x)\left( \frac{1}{\cos x} \right)\]
Getting it so far?

Answer 33

yes po .. you separate the two ..

Answer 34

Well, we're going to separate a little more, in particular, let's express cot x as
cos x / sin x
shall we?
\[=\left(\frac{\cos x}{\sin x}\right)\left(\frac{\cos x}{\sin x}\right)\left(\frac{1}{\cos x}\right)\]
Saw how I did that?
Remember that it was cot² x, originally...

Answer 35

aha! i get it. :D you separate cot^2 x in a simplest form ..

Answer 36

Actually, that wasn't necessary... I can replace one of them with cot x again.
cos x / sin x = cot x, right?
Well
\[=(\cot x)\left(\frac{\cos x}{\sin x}\right)\left(\frac{1}{\cos x}\right)\]
okay?

Answer 37

aha! you're right. then po?

Answer 38

LOL
Anyway...
You can see that you can cancel out the cos x in the expression, right?

Answer 39

yes po. hehe

Answer 40

Well, cancel out and you get
\[(\cot x)\left( \frac{1}{\sin x}\right)\]
Can you see it now?
What's 1/sin x
?

Answer 41

csc x po. haha. i know the answer now. thank you po. :D

Answer 42

heh