Question

more identity troubles.......

Answer 1

\[\frac{ \cot^2x }{ cscx+1 }=\frac{ 1-sinx }{ sinx }\]

Answer 2

verify the identity

Answer 3

did you change everything to sin and cos ?

Answer 4

\[\frac{ \frac{ \cos }{ \sin }^{2} }{ \frac{ 1 }{ \sin }+1 }\]

Answer 5

yes along as the top is cos^2/sin^2

Answer 6

I would add the bottom to get (1+sin)/sin
then invert and multiply

Answer 7

and because you are looking for sin (on the right side)
I would change cos^2 to 1-sin^2

Answer 8

how far did you get ?

Answer 9

\[\frac{ \frac{ 1-\sin^2 }{ \sin } }{ \frac{ 1+\sin }{ \sin } }\]

Answer 10

is that correct

Answer 11

almost. it should say
\[ \frac{ \frac{ 1-\sin^2 }{ \sin^2 } }{ \frac{ 1+\sin }{ \sin } } \]

Answer 12

now change the division by (1+sin)/sin to a multiply by sin/(1+sin)

Answer 13

remember
\[ \cot^2 = \frac{\cos^2}{\sin^2} \]

Answer 14

OH! the top changes to cos^2/sin^2 because 1-sin^2=cos^2!

Answer 15

we started with cot^2 change to cos^2/sin^2 and then to (1-sin^2)/sin^2

that is the top
now multiply by the inverse of the bottom.
multiply by sin/(1+sin)

Answer 16

I would factor the 1-sin^2 (it's a difference of squares)

Answer 17

Oh... so when you cancel out you get 1-sin^2/sin

Answer 18

how about this:
\[ \frac{ \frac{ 1-\sin^2 }{ \sin^2 } }{ \frac{ 1+\sin }{ \sin } } = \frac{ 1-\sin^2 }{ \sin^2 } \cdot \frac{ \sin }{ 1+\sin }\]

Answer 19

or I mean 1-sin/sin

Answer 20

you can factor 1-sin^2 to (1-sin)(1+sin)

Answer 21

and then