Question

Establish the Identity csc u - cot u = (sin u)/(1+cos u)

Answer 1

@mathmale

Answer 2

cscu=1/sinu

Answer 3

and
cot u =cosu/sinu

Answer 4

cscu-cotu=1/sinu -cosu/sinu

Answer 5

=1-cosu/sinu

Answer 6

Hold up esamalaa, I'm trying to work it out given what you've told me

Answer 7

and we are subtracting cot u from csc u. Using esamalaa's equivalencies, \[\frac{ 1 }{ \sin u }-\frac{ \cos u }{ \sin u },\], which is relatively straightforward to simplify because we have the same denominator in both fractions. @noseboy, please perform the indicated subtraction.

Answer 8

That I shall. "1/sin - cos/sin" = "(1-cos)/sin"

Answer 9

Is that equal to the right side of your original equation?

Answer 10

And 1-cos=sin, correct?

Answer 11

then you now have
\[\frac{ 1-cosu }{ sinu }=\frac{ sinu }{ 1+cosu }\]

Answer 12

Yes indeed @esamalaa

Answer 13

\[\sin ^{2}u=(1-cosu)(1+cosu)\]

Answer 14

Where did the squared come from?

Answer 15

\[\sin ^{2}u=\cos ^{2}u-1\]

Answer 16

@esamalaa: Wouldn't it be \[\sin ^{2}\theta=1-\cos ^{2}\theta ?\]

Answer 17

yes i am so sorry

Answer 18

Would you mind if we start from here and work forward?\[\frac{ 1 }{ \sin u }-\frac{ \cos u }{ \sin u }?\]

Answer 19

How would you go about showing that this equals the right side, \[\frac{ \sin u }{ 1+cosu }\]

Answer 20

?

Answer 21

Hint: Combine the two fractions with sin u in their denominators. We get \[\frac{ 1-\cos u }{ \sin u }\]

Answer 22

What next?

Answer 23

hello
\[\frac{ 1-cosu }{ sinu }\times \frac{ 1+cosu }{ 1+cosu }\rightarrow \frac{ 1-\cos ^{2}u }{ sinu+sinu cosu }\rightarrow \frac{ \sin ^{2}u }{ sinu+sinu cosu }\rightarrow \frac{ sinu }{ 1+cosu }\]

Answer 24

@mathmale

Answer 25

Great work, esamalaa, and i appreciate the way you show all your work so clearly. Thanks. See you again later on! MM

Answer 26

Thank you both very much!