The problem is..... (tan[theta] + cot[theta])^2=sec^2[theta] + scs^2[theta]

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- May 22nd 2006, 12:17 PMkennysbaby28Analytic Trig Identities Problem. Need help asap!
The problem is..... (tan[theta] + cot[theta])^2=sec^2[theta] + scs^2[theta]

- May 22nd 2006, 12:42 PMsweetQuote:

Originally Posted by**kennysbaby28**

solved....

(tan[theta] + cot[theta])^2=

=(tan[theta])^2 +2(tan[theta]) ( cot[theta]) +(cot[theta])^2

=(tan[theta])^2 +2(sin[theta]/cos[theta])(cos[theta]/sine[theta]+(cot[theta])^2

=(tan[theta])^2 +2+(cot[theta])^2

={(tan[theta])^2 +1}+{(cot[theta])^2+1}

an we know

(tan[theta])^2 +1=sec^2[theta] and

(cot[theta])^2 +1=csc^2[theta]

then

(tan[theta] + cot[theta])^2=sec^2[theta] + csc^2[theta] - May 22nd 2006, 01:29 PMkennysbaby28Question
I'm not sure on how you got what you did and I'm not doubting you. It's just hard to understand the way it's written out. you have two sides to the problem so what goes on what side? Thanks for helping me!

- May 22nd 2006, 01:50 PMCaptainBlackQuote:

Originally Posted by**kennysbaby28**

RonL - May 22nd 2006, 01:55 PMsweetQuote:

Originally Posted by**kennysbaby28**

u take R.H.s

and know u have sin[theta] in numerator and sin[theta] in denominator also u have cos[theta] in numerator and cos[theta] in denominator

then they drop so the eq be come

and we know

and

then

- May 28th 2006, 01:20 PMSoroban
Hello, kennysbaby28!

Quote:

Since , we have:

Since and

. . we have: - May 29th 2006, 11:18 PMmalaygoel
i will use tanx and cotx

firstly, change tanx and cotx in sinx and cosx.

add both of them

you will get 1/(sinx*cosx)

square the expression

change 1 into sinx^2 +cosx^2

you got the answer - Jun 4th 2006, 02:37 AMchancey