(tan(theta)-cot(theta))/(tan(theta)+cot(theta) = 1-cos^2(the

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camgrover

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Apr 10, 2008
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Ok, I've been stuck for like 3 hours. This seems impossible:

(tan(theta)-cot(theta))/(tan(theta)+cot(theta) = 1-cos^2(theta)

Help...please
 
Did you know that cot(theta) = 1 / tan(theta) ?

And that 1 - cos^2(theta) = sin^2(theta) ?
 
Re: Seemingly Impossible trig identitiy

Hello, camgrover!

Did you copy it wrong? . . . It is not an identity.

tan⁡θ−cot⁡θtan⁡θ+cot⁡θ = 1−cos⁡2 ⁣θ⏟???\displaystyle \frac{\tan\theta - \cot\theta}{\tan\theta+\cot\theta} \:= \:\underbrace{1-\cos^2\!\theta}_{???}
Click to expand...

The left side is:  sin⁡θcos⁡θ−cos⁡θsin⁡θsin⁡θcos⁡θ+cos⁡θsin⁡θ\displaystyle \text{The left side is: }\:\frac{\dfrac{\sin\theta}{\cos\theta} - \dfrac{\cos\theta}{\sin\theta}}{\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}}


Multiply top and bottom by sin⁡θcos⁡θ ⁣:    sin⁡2 ⁣θ−cos⁡2 ⁣θsin⁡2 ⁣θ+cos⁡2 ⁣θ  =  −(cos⁡2 ⁣θ−sin⁡2 ⁣θ)1\displaystyle \text{Multiply top and bottom by }\sin\theta\cos\theta\!:\;\;\frac{\sin^2\!\theta - \cos^2\!\theta}{\sin^2\!\theta + \cos^2\!\theta} \;=\; \frac{-(\cos^2\!\theta-\sin^2\!\theta)}{1}


Therefore, we have:   −cos⁡2θ\displaystyle \text{Therefore, we have: }\;-\cos2\theta

 
camgrover said:
Ok, I've been stuck for like 3 hours. This seems impossible:

(tan(theta)-cot(theta))/(tan(theta)+cot(theta) = 1-cos^2(theta)

Help...please
Click to expand...
are you sure it is not:


(tan(theta)-cot(theta))/(tan(theta)+cot(theta) = 1 - 2 * cos^2(theta)
 
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