Simplify the Trig Expression...

**deezeejoey** · February 6th 2008, 09:38 AM

I'm stuck and i don't know how to do this.

(4tan^2(θ)cot^2(θ)/csc^2(θ))+(4/tan^2(θ)+1)

**colby2152** · February 6th 2008, 09:47 AM

Originally Posted by deezeejoey

I'm stuck and i don't know how to do this.

(4tan^2(θ)cot^2(θ)/csc^2(θ))+(4/tan^2(θ)+1)

Did you try reducing everything to sines and cosines?

Hint: The second term would then be: $\frac{4cos^2(\theta)}{sin^2(\theta)+cos^2(\theta)} = 4cos^2(\theta)$

**Soroban** · February 6th 2008, 12:31 PM

Hello, deezeejoey!

Do you remember any of the basic identities?

$\frac{4\tan^2\!\theta\cot^2\!\theta}{\csc^2\!\thet a}+ \frac{4}{\tan^2\!\theta+1}$

$\text{The first fraction is: }\;\frac{4(\overbrace{\tan\theta\cot\theta}^{\text {This is 1}})^2}{\csc^2\!\theta} \;=\;\frac{4}{\csc^2\!\theta} \;=\;4\sin^2\!\theta$

$\text{The second fraction is: }\;\frac{4}{\underbrace{\tan^2\!\theta + 1}_{\text{This is }\sec^2\theta}} \;=\;\frac{4}{\sec^2\!\theta} \;=\;4\cos^2\!\theta$

$\text{The problem becomes: }\;4\sin^2\!\theta + 4\cos^2\!\theta \;=\;4\underbrace{(\sin^2\!\theta + \cos^2\!\theta)}_{\text{This is 1}} \;=\;4\cdot1 \;=\;4$

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