# Thread: Analytic Trig Identities Problem. Need help asap!

1. ## Analytic Trig Identities Problem. Need help asap!

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

2. 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]

3. ## 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!

4. 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

5. 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]
$

6. Hello, kennysbaby28!

$(\tan\theta + \cot\theta)^2 \:= \:\sec^2\theta + \csc^2\theta$
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: $(\sec^2\theta - 1) + 2 + (\csc^2\theta - 1) \;= \;\sec^2\theta + \csc^2\theta$

7. 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

8. $(\tan \theta + \cot \theta)^2 = \sec^2 \theta + \csc^2 \theta$

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

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

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

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

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

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

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

$(1)^2 = 1$