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?

Mathematics  Trigonometric Identities 
Reiny,
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,
I'm not suppose to use the inverse identities yet ...

Mathematics  Trigonometric Identities 
Reiny,
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,
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,
yes
correct 
Mathematics  Trigonometric Identities 
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. 
Mathematics  Trigonometric Identities 
Janel,
cscxcotx forms an identity with?