Mathematics  Trigonometric Identities
posted by Anonymous on .
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
= RS 
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:
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 
yes
correct 
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. 
cscxcotx forms an identity with?