Question

Establish the identity.

Answer 1

\[\sec u + \tan u=\frac{ \cos u }{ 1-\sin u }\]

Answer 2

@Hero @Directrix

Answer 3

did you try writing left hand side in terms of sin(u) and cos(u)

Answer 4

and then combining fractions

Answer 5

Yes i got stuck

Answer 6

and recall 1-sin^2(u) is cos^2(u)
so after the last step I mention multiply top and bottom by 1-sin(u)

Answer 7

please let me know or show how far you have gotten

Answer 8

so i changed to

\[\frac{ 1+\sin u }{ \cos u }\]

Answer 9

I was going to say multiply 1+sinu on top and bottom ?

Answer 10

well you want to multiply top and bottom by 1-sin(u) since (1+sin(u))(1-sin(u)) is 1-sin^2(u) which is cos^2(u)

Answer 11

Oh yea you switch the signs when you do that i forgot. Hold on

Answer 12

I basically went in a circle and got 1-sinu

Answer 13

not sure what you mean

Answer 14

I multiplied all around and got 1-sinu

Answer 15

\[\frac{1+\sin(u)}{\cos(u)} \cdot \color{red}{\frac{1-\sin(u)}{1-\sin(u)} } \\ \frac{\cos^2(u)}{\cos(u)(1-\sin(u))}\]
you should see something that cancels

Answer 16

wait 1-sin^2u i cos^2u?

Answer 17

@myininaya

Answer 18

yes?

Answer 19

Why is 1-sin^2 = to cos^2u ?

Answer 20

have you guys mentioned the identity sin^2(u)+cos^2(u)=1 "?

Answer 21

i can explain that identity if needed
it starts off with me drawing a unit circle and a right triangle in the first quadrant

Answer 22

Yea that i know but thats sin^2 + cos^2

Answer 23

you mean sin^2(u)+cos^2(u)=1 right?

Answer 24

so we know sin^2(u)+cos^2(u)=1
so subtracting sin^2(u) on both sides of the equation will mean the equation still holds

Answer 25

But when you did
1+sin^2u X 1-sin^2u = cos^2

Answer 26

\[\sin^2(u)+\cos^2(u)=1\]
\[\sin^2(u)+\cos^2(u)-\color{red}{\sin^2(u)}=1-\color{red}{\sin^2(u)}\]
\[\cos^2(u)=1-\sin^2(u)\]

Answer 27

Oh i remember now but when you do that doesn't i have to be

1-cos^2u ?

Answer 28

and i didn't do 1+sin^2(u)X1-sin^2(u)
I did (1+sin(u))(1-sin(u))

Answer 29

no 1-sin^2(u) isn't 1-cos^2(u)

Answer 30

the identity is that sin^2(u)+cos^2(u)=1
which means 1-sin^2(u) can be written as cos^2(u) not 1-cos^2(u)

Answer 31

are you trying to figure out how (1+sin(u))(1-sin(u)) is 1-sin^2(u)

or

are you trying to figure out how 1-sin^2(u) is cos^2(u)

Answer 32

when multiplying conjugates you only have to multiply first and last
you and do inner and outer though but this will cancel out
\[
(1+u)(1-u) \\
1-u+u-u^2 \\
1-u^2\]

Answer 33

\[(1+\sin(u))(1-\sin(u)) \\ 1-\sin(u)+\sin(u)-\sin^2(u) \\ 1-\sin^2(u) \text{ since } -\sin(u)+\sin(u)=0\]

Answer 34

No i got it now ok that part is ok. So that means

1-cos^2u = sin^2?

And the bottom part i i actually multiplied sin and cos and was freaking me out cause i couldn't get rid of the cos^2u on top but now i see you don't multiply them :D

Answer 35

yeah you can cancel out a factor of cos(u) from top and bottom

Answer 36

\[\frac{1+\sin(u)}{\cos(u)} \cdot \color{red}{\frac{1-\sin(u)}{1-\sin(u)} } \\ \frac{\cos^2(u)}{\cos(u)(1-\sin(u))} \\ \frac{\cos(u) \cos(u)}{\cos(u)(1-\sin(u)} \\ \frac{\cancel{\cos(u) } \cos(u)}{\cancel{\cos(u)}(1-\sin(u))}\]

Answer 37

OHH ok thank you :D

Answer 38

np