Question

prove identity: (1+sinx)/(1-sinx)=([1/sinx]+1)/([1/sinx]-1)

Answer 1

starting with right hand side

1
-- + 1 = 1 + sin x
sin x --------
sin x

1
-- - 1 = 1 - sinx
sin x -------
sin x

dividng first exoression by the last, we invert the last one and multiply:

so this is

1 + sin x * sin x
-------- ------
sin x 1 - sin x

= (1 + sin x ) / ( 1 - sin x)

Answer 2

...Ohh i got lost. Can you do it step by step please?

Answer 3

the rhs is 1/sinx+1/1/sinx-1, so it becomes 1+sinx/sinx?

Answer 4

no (1/sinx )+ 1 = (1 + sinx) / sinx

and ( 1 /sinx) - 1 = (1 - sinx) / sin x

Answer 5

and dividing the 2 above results
gives the left side

Answer 6

OHHHH OMGOSH I GET IT!

Answer 7

if the two fractions were 3 / 4 and 4 / 5

then i to work out 3/4 / 4/5
we change 4/5 to 5/4 and then multiply

3/4 * 5/4 = 15/16

if the numbers are replaced by terms in x or sinx etc
the same rules apply

Answer 8

ohs! i see

Answer 9

good!!

Answer 10

can you help me with another question..i kind of got stuck = u =

Answer 11

[(1/cosx)+tanx](1-sinx)=cosx

i know i have to work on the LHS and i don't know if i should put tanx to sinx/cosx or change 1/cosx to secx