Question

Can someone check my work for a problem?

Answer 1

\[\frac{ 1 }{ 1 + cosB } + \frac{ 1 }{ 1 - cosB } \rightarrow \frac{ 1 }{ 1 - \cos^2B } \rightarrow \frac{ 2 }{ \sin^2B } \rightarrow 2\csc ^2 B\]

Answer 2

@Nnesha Could you check this for me? If you don't mind.

Answer 3

well well 2nd step isn't correct how did you get 2 at the numerator ?? u forgot one step

Answer 4

I just put it since it was squared.

Answer 5

you forgot the same steps like on the previous post :=))
when we find common denominator we should multiply the top and bottom of the fraction

Answer 6

remember denominators aren't the same

Answer 7

Oh!

\[\frac{ (1 + cosB) (1 - cosB) }{ 1 - \cos^2B }\]

Answer 8

Cross out 1 - cosB on top, and one on bottom, so

\[\frac{ 1 + cosB }{ 1 - cosB }\]

Answer 9

ye but there supposed to be plus sign \[\frac{ 1 }{ 1 + cosB }\color{Red}{ + }\frac{ 1 }{ 1 - cosB } \]


\[\frac{ (1 + cosB)\color{Red}{+} (1 - cosB) }{ 1 - \cos^2B }\] and then combine like terms

Answer 10

and since there is a plus sign just remove the parentheses

Answer 11

\[\frac{ 2 - cosB }{ 1 - \cos^2B }\]

Answer 12

Which turns into sin^2x

Answer 13

Oh wait, I forgot I have another ten minutes. Lol.

Answer 14

remember find common denominators when the denominators aren't the same
we should multiply the numerator of first fraction by the denominator of 2nd fraction
and multiply numerator of 2nd fraction by the denominator of first fraction