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

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

- Mathematics - Trigonometric Identities -
**Reiny**, Thursday, November 8, 2007 at 6:49pm
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**, Thursday, November 8, 2007 at 6:52pm
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 -
**Reiny**, Thursday, November 8, 2007 at 7:44pm
yes

correct

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

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