# Analytic Trig Identities Problem. Need help asap!

• May 22nd 2006, 11:17 AM
kennysbaby28
Analytic Trig Identities Problem. Need help asap!
The problem is..... (tan[theta] + cot[theta])^2=sec^2[theta] + scs^2[theta]
• May 22nd 2006, 11:42 AM
sweet
Quote:

Originally Posted by kennysbaby28
The problem is..... (tan[theta] + cot[theta])^2=sec^2[theta] + scs^2[theta]

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, 12:29 PM
kennysbaby28
Question
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, 12:50 PM
CaptainBlack
Quote:

Originally Posted by kennysbaby28
The problem is..... (tan[theta] + cot[theta])^2=sec^2[theta] + scs^2[theta]

Please do not post the same question twice in different fora.

RonL
• May 22nd 2006, 12:55 PM
sweet
Quote:

Originally Posted by kennysbaby28
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!

no prblem
u take R.H.s
$
(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
$

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

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

and 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 28th 2006, 12:20 PM
Soroban
Hello, kennysbaby28!

Quote:

$(\tan\theta + \cot\theta)^2 \:= \:\sec^2\theta + \csc^2\theta$
Expand the left side: $\tan^2\theta + 2\tan\theta\cot\theta + \cot^2\theta$

Since $\tan\theta\cdot\cot\theta\:=\:1$, we have: $\tan^2\theta + 2 + \cot^2\theta$

Since $\tan^2\theta\:=\:\sec^2\theta - 1$ and $\cot^2\theta \:= \:\csc^2\theta - 1$

. . we have: $(\sec^2\theta - 1) + 2 + (\csc^2\theta - 1) \;= \;\sec^2\theta + \csc^2\theta$
• May 29th 2006, 10:18 PM
malaygoel
i will use tanx and cotx
firstly, change tanx and cotx in sinx and cosx.
you will get 1/(sinx*cosx)
square the expression
change 1 into sinx^2 +cosx^2
• June 4th 2006, 01:37 AM
chancey
$(\tan \theta + \cot \theta)^2 = \sec^2 \theta + \csc^2 \theta$

$\tan^2 \theta + 2 \tan \theta \cot \theta + \cot^2 \theta = \sec^2 \theta + \csc^2 \theta$

$\tan^2 \theta + 2 + \cot^2 \theta = \sec^2 \theta + \csc^2 \theta$

$\frac{\sin^2 \theta}{\cos^2 \theta} + 2 + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}$

$\frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} + 2 = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}$

$\frac{2 \sin^2 \theta \cos^2 \theta + \sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta}$

$2 \sin^2 \theta \cos^2 \theta + \sin^4 \theta + \cos^4 \theta = 1$

$(\sin^2 \theta + \cos^2 \theta)^2 = 1$

$(1)^2 = 1$