Tuesday
March 28, 2017

Post a New Question

Posted by 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 - ,

    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 - ,

    I'm not suppose to use the inverse identities yet ...

  • Mathematics - Trigonometric Identities - ,

    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 - ,

    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

  • Mathematics - Trigonometric Identities - ,

    yes
    correct

  • Mathematics - Trigonometric Identities - ,

    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 - ,

    cscx-cotx forms an identity with?

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question