Verify Trig Identities

**RosieLaird** · August 5th 2010, 02:31 PM

(CosθCotθ)/1-Sinθ - 1 = Cscθ

4(Tan^4)x + (tan^2)x - 3 = (sec^2)x[4(tan^2)x - 3]

Sin(α/3)Cos(α/3) = .5Sin(2α/3)

**skeeter** · August 5th 2010, 02:47 PM

Originally Posted by RosieLaird

(CosθCotθ)/1 = Cscθ

4(Tan^4)x + (tan^2)x - 3 = (sec^2)x[4(tan^2)x - 3]

Sin(α/3)Cos(α/3) = .5Sin(2α/3)

1. recheck the first equation ... as you typed it, it makes no sense.

2. factor the left side ... then use the identity $1+\tan^2{x} = \sec^2{x}$

3. use double angle formula for sine on the right side

**Soroban** · August 6th 2010, 10:52 AM

Hello, RosieLaird!

$\dfrac{\cos\theta \cot\theta}{1-\sin\theta} - 1 \:=\: \csc\theta$

The left side is: . $\dfrac{\cos\theta\cdot\frac{\cos\theta}{\sin\theta }}{1-\sin\theta} - 1 \;\;=\;\;\dfrac{\cos^2\!\theta}{\sin\theta(1-\sin\theta)} - 1$

. . $=\;\;\dfrac{\cos^2\theta - \sin\theta(1 - \sin\theta)}{\sin\theta(1-\sin\theta)} \;\;=\;\;\dfrac{\overbrace{\cos^2\theta + \sin^2\theta}^{\text{This is 1}} - \sin\theta}{\sin\theta(1-\sin\theta)}$

. . $=\;\;\dfrac{1-\sin\theta}{\sin\theta(1-\sin\theta)} \;\;=\;\;\dfrac{1}{\sin\theta} \;\;=\;\;\csc\theta$

**sa-ri-ga-ma** · August 6th 2010, 08:37 PM

4(Tan^4)x + (tan^2)x - 3 = (sec^2)x[4(tan^2)x - 3]

$4tan^2(x)tan^2(x) + tan^2(x) - 3$

$4tan^2(x)(sec^2x-1) + tan^2(x) - 3$

$4tan^2(x)sex^2(x) - 4tan^2(x) + tan^2(x) -3$

$4tan^2(x)sex^2(x) - 3tan^2(x) -3$

$4tan^2(x)sex^2(x) - 3(tan^2(x) + 1)$

$4tan^2(x)sex^2(x) - 3sec^2(x)$

$sec^2(x)[4tan^2(x) - 3]$

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