1. ## Verify Trig Identities

(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)

2. 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 $\displaystyle 1+\tan^2{x} = \sec^2{x}$

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

3. Hello, RosieLaird!

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

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

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

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

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

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

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

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

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

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

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

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