you confused me too
I will start again with
LS = 1 + cot^2 y
= 1 + cos^2 y/sin^2 y , now find a common denominator
= (sin^2 y + cos^2 y) / sin^2 y, but sin^2 y + cos^2 y=1
= 1/sin^2 y
I'm not suppose to use the inverse identities yet ...
ok, then in
LS = 1 + 1/tan^2y
= 1 + 1/(sin^2 / cos^2 y)
= 1 + (cos^2 y)/(sin^2 y)
= ... my second line
surely you are going to use tanx = sinx/cosx !!!
we are not using "inverse identities" here
I meant the reciprocals of SOH CAH TOA
And ... I just got it
Here's my answer:
= 1 + 1/tan^2y
= 1 + 1/(sin^2y/cos^2y)
= 1 + 1(cos^2y/sin^2y)
= 1 + cos^2y / sin^2y
= 1 + 1-sin^2y / sin^2y
= 1(sin^2y) / sin^2y + 1 - sin^2y / sin^2y
= sin^2y + 1 - sin^2y / sin^2y
should have taken a closer look at your solution.
the last few lines make no sense with "no brackets" being used
why don't you follow the steps of my original solution, it is so straightforward.
cscx-cotx forms an identity with?
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