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March 1, 2015

Posted by **Anonymous** on Thursday, November 8, 2007 at 6:17pm.

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**, Thursday, November 8, 2007 at 6:33pmyou 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**, Thursday, November 8, 2007 at 6:37pmI'm not suppose to use the inverse identities yet ...

- Mathematics - Trigonometric Identities -
**Reiny**, Thursday, November 8, 2007 at 6:49pmok, 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**, Thursday, November 8, 2007 at 6:52pmI 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 -
**Reiny**, Thursday, November 8, 2007 at 7:44pmyes

correct

- Mathematics - Trigonometric Identities -
**Reiny**, Thursday, November 8, 2007 at 8:13pmshould 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 -

- Mathematics - Trigonometric Identities -
- Mathematics - Trigonometric Identities -
**Janel**, Wednesday, April 9, 2008 at 3:26pmcscx-cotx forms an identity with?

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