Question

Rewrite the expression so that there is no denominator:
(tan2x)/(cscx+1)

Answer 1

do you mean tan 2x or tan ^2 x ?

Answer 2

tan 2x :)

Answer 3

\[\frac{ \tan 2x }{ \csc x+1 }=\frac{ \tan 2x }{ \frac{ 1 }{\sin x }+1 }=\]

Answer 4

\[\frac{ \tan 2x }{ \frac{ 1 }{\sin x }+1 }=\frac{ \tan 2x }{ \frac{ 1+sinx }{\sin x }}=\]

Answer 5

did you try it ?

Answer 6

yea :)

Answer 7

and what was your problem ?

Answer 8

I'm not sure how to go on from there to get rid of the denominator...

Answer 9

show me your work

Answer 10

I stopped there, and I don't know how to get rid of the denominator...

Answer 11

\[\frac{ \tan 2x }{ \csc x+1 }\times \frac{ \csc x-1 }{ \csc x-1 }= \frac{ \tan 2x (\csc x-1) }{ \csc^2 x-1 }=\]

Answer 12

\[\csc ^2 x -1=\cot ^2 x\]

Answer 13

do you fin out
what would you do ?

Answer 14

\[ \frac{ \tan 2x (\csc x-1) }{ \cot^2 x }=\tan 2x (\csc x-1) \tan^2 x\]

Answer 15

Thank you so much! :)