Mathematics  Trigonometric Identities
Let y represent theta
Prove:
1 + 1/tan^2y = 1/sin^2y
My Answer:
LS:
= 1 + 1/tan^2y
= (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y)
= (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y)
= (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y)
= (sin^2y + cos^2y) + (sin^2y + cos^2y)(sin^2y) / (sin^2y)(cos^2y/sin^2y)
... And now I confuse myself
Where did I go wrong? And please direct me on how to fix it?

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
= RSposted by Reiny

I'm not suppose to use the inverse identities yet ...
posted by Anonymous

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
posted by Reiny

I meant the reciprocals of SOH CAH TOA
And ... I just got it
Here's my answer:
LS:
= 1 + 1/tan^2y
= 1 + 1/(sin^2y/cos^2y)
= 1 + 1(cos^2y/sin^2y)
= 1 + cos^2y / sin^2y
= 1 + 1sin^2y / sin^2y
= 1(sin^2y) / sin^2y + 1  sin^2y / sin^2y
= sin^2y + 1  sin^2y / sin^2y
= 1/sin^2yposted by Anonymous

yes
correctposted by Reiny

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.posted by Reiny

cscxcotx forms an identity with?
posted by Janel
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