Question

Simplify the expression.

\(\dfrac{(csc~y~+~cot~y)(csc~y~-~cot~y)}{csc~y}\)

-sin y
cos y
sin y \(\leftarrow\) I think this one
csc y

Answer 1

(a+b)(a-b)=a^2-b^2

Answer 2

and i remember that sin^2(x)+cos^2(x)=1
so 1+cot^2(x)=csc^2(x)
which implies 1=csc^2(x)-cot^2(x)

Answer 3

you are right

Answer 4

So it would be \(\dfrac{csc^2y-cot^2y}{cscy}\)?

Answer 5

yeah

Answer 6

then apply the pythagorean identity i just mentioned

Answer 7

are however you spell that one guy's name

Answer 8

then \(\dfrac{1}{cscy}=siny\)

Answer 9

yep yep

Answer 10

Thanks! @myininaya and @satellite73