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?

asked by Anonymous
  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

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

    posted by Anonymous
  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


    posted by Reiny
  4. 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 + 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

    posted by Anonymous
  5. yes
    correct

    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.

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

    posted by Janel

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