# Mathematics - Trigonometric Identities

Let y represent theta

Prove:

1 + 1/tan^2y = 1/sin^2y

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?

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3. 👁 135
1. 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

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

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3. 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

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posted by Reiny
4. I meant the reciprocals of SOH CAH TOA

And ... I just got it

LS:

= 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
= 1/sin^2y

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5. yes
correct

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posted by Reiny
6. 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.

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posted by Reiny
7. cscx-cotx forms an identity with?

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posted by Janel

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