Question

Trigonometry help?

Answer 1

@Nnesha \[\frac{ \cos \theta }{ 1-\sin \theta } + \frac{ 1-\sin \theta }{ \cos \theta }\]

Answer 2

i love that type of question
that is equal to what there should be an equal sign

Answer 3

sorry, it's just simplify :)

Answer 4

find common denominator

Answer 5

\[\cos \theta(1-\sin \theta)\]

Answer 6

right when we find common denominator
we should multiply the numerator of 1st fraction by the denominator of 2nd fraction
multiply numerator of 2nd fraction by the denominator of 1st fraction \[\frac{ \cos \theta }{\color{Red}{ 1-\sin \theta} } + \frac{ 1-\sin \theta }{\color{blue}{ \cos \theta} }\]

\[\frac{ \color{blue}{\cos \theta}*\cos \theta + \color{Red}{(1-\sin \theta)}(1-\sin \theta) }{ \cos \theta(1-\sin \theta) }\]

Answer 7

simplify that.

Answer 8

wouldn't that be \[\cos \theta + (1-\sin \theta)\]

Answer 9

I don't know how else to simplify that...

Answer 10

did you get this cos theta+ (1-sinx) after simplifying ?
and
are you sure it's (1 `-` sinx) at the numerator and denominator
im just guessing one of them supposed to be positive (1+sinx)

Answer 11

one of the cos thetas and one of the (1-sin theta)s should cancel, right? That leaves cos theta + (1-sin theta), doesn't it?

Answer 12

No0o0o!!! you can't do that

Answer 13

\[\frac{ \color{blue}{\cancel{\cos \theta}}*\cos \theta + \color{Red}{\cancel{(1-\sin \theta)}}(1-\sin \theta) }{ \cancel{\cos} \theta\cancel{(1-\sin \theta)} }\]
This is NOT allowed.
bec that's same as
\[\rm \frac{ \cos \theta*\cos \theta }{ \cos(1-\sin \theta) } +\frac{(1-\sin \theta)(1-\sin \theta)}{\cos \theta(1-\sin \theta)}\]

Answer 14

both term at numerator are dividing by cos theta (1-sin theta)

Answer 15

that' would be the next step

Answer 16

make sense ?? :=))

Answer 17

\[\frac{ \cos^2\theta+\sin^2\theta-2\sin \theta + 1 }{ \cos \theta(1-\sin \theta) }\]

Answer 18

looks good now use the identity \[\rm \cos^2 \theta + \sin^2 \theta =1 \]
solve this for cos^2

Answer 19

\[\frac{ \color{Red}{\cos^2\theta}+\sin^2\theta-2\sin \theta + 1 }{ \cos \theta(1-\sin \theta) }\]
cos^2 theta = ??? use the identity

Answer 20

cos^2=1-sin^2

Answer 21

right replace cos^2 x with 1-sin x^2 and then simplify

Answer 22

\[\frac{ \color{Red}{1 - sin^2 \theta}+\sin^2\theta-2\sin \theta + 1 }{ \cos \theta(1-\sin \theta) }\]

Answer 23

\[\frac { 1-\sin \theta + \sin^2\theta - 2\sin \theta + 1 } { \cos \theta(1-\sin \theta) }\]

Answer 24

sin^2 ** u forgot the 2 :=)) now try to simplify that

Answer 25

can the 1-sin theta cancel out?

Answer 26

what 1-sintheta are you talking about ?
what did you get at the numerator ? :=))

Answer 27

would it be this? I'm so confused... \[\frac {\sin^2\theta-2\sin \theta + 1} { \cos \theta }\]

Answer 28

well cos^2 theta = 1-cos^2 so substitute \[\frac{ \color{Red}{1 - sin^2 \theta}+\sin^2\theta-2\sin \theta + 1 }{ \cos \theta(1-\sin \theta) }\]
now simplify the numerator first

Answer 29

combine like terms!!

Answer 30

1-sin^2x+sin^2x -2sinx +1 what are like terms ?

Answer 31

sin^2x

Answer 32

so it's 2-2sinx?

Answer 33

right !! now take out the common factor from the numerator \[\frac{ 2 - 2\sin \theta }{ \cos (1-\sin theta) }\]

Answer 34

2/cos theta! or 2sec theta! (My professor wants it in terms of sec)

Answer 35

alright then 2 sec theta !!

Answer 36

Thank you so much! :D

Answer 37

np , great work!!