Posted by Anonymous on Thursday, November 8, 2007 at 6:17pm.
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?

Mathematics  Trigonometric Identities  Reiny, Thursday, November 8, 2007 at 6:33pm
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
= RS

Mathematics  Trigonometric Identities  Anonymous, Thursday, November 8, 2007 at 6:37pm
I'm not suppose to use the inverse identities yet ...

Mathematics  Trigonometric Identities  Reiny, Thursday, November 8, 2007 at 6:49pm
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

Mathematics  Trigonometric Identities  Anonymous, Thursday, November 8, 2007 at 6:52pm
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^2y

Mathematics  Trigonometric Identities  Reiny, Thursday, November 8, 2007 at 7:44pm
yes
correct

Mathematics  Trigonometric Identities  Janel, Wednesday, April 9, 2008 at 3:26pm
cscxcotx forms an identity with?
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