# Question #45dfa

Apr 26, 2016

The 3 first expressions are lacking some parentheses. I'm not sure if I put the missing parentheses in the right places, but in my answer the identities hold true.

#### Explanation:

1) $\frac{1 - {\cos}^{2} \theta}{{\sec}^{2} \theta - {\tan}^{2} \theta + \cos \theta} + \cos \theta = 1$

$\frac{1 - {\cos}^{2} \theta}{\frac{1}{\cos} ^ 2 \theta - {\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta + \cos \theta} + \cos \theta = 1$
$\frac{1 - {\cos}^{2} \theta}{\frac{1 - {\sin}^{2} \theta}{\cos} ^ 2 \theta + \cos \theta} + \cos \theta = 1$
$\frac{1 - {\cos}^{2} \theta}{\frac{\cancel{{\cos}^{2} \theta}}{\cancel{{\cos}^{2} \theta}} + \cos \theta} + \cos \theta = 1$
$\frac{\left(1 - \cos \theta\right) \cancel{\left(1 + \cos \theta\right)}}{\cancel{1 + \cos \theta}} + \cos \theta = 1$
$1 - \cancel{\cos \theta} + \cancel{\cos \theta} = 1$
$1 = 1$ (That's true and therefore the original identity holds true)

2) $\frac{{\sin}^{2} x + 4 \sin x + 3}{\cos} ^ 2 x = \frac{\sin x + 3}{1 - \sin x}$

$\frac{{\sin}^{2} x + 4 \sin x + 3}{\cancel{{\cos}^{2} x}} = \frac{\left(\sin x + 3\right) \left(1 + \sin x\right)}{\cancel{\left(1 - \sin x\right) \left(1 + \sin x\right)}}$
${\sin}^{2} x + 4 \sin x + 3 = \sin x + {\sin}^{2} x + 3 + 3 \sin x$
${\sin}^{2} x + 4 \sin x + 3 = {\sin}^{2} x + 4 \sin x + 3$ (That's true and therefore the original identity holds true)

3) $\frac{\sec \theta \cdot \sin \theta}{\tan \theta + \cot \theta} = {\sin}^{2} \theta$

$\frac{1}{\cos} \theta \cdot \sin \frac{\theta}{\sin \frac{\theta}{\cos} \theta + \cos \frac{\theta}{\sin} \theta} = {\sin}^{2} \theta$
$\frac{1}{\cancel{\cos \theta}} \cdot \sin \frac{\theta}{\frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\cancel{\cos \theta} \sin \theta}} = {\sin}^{2} \theta$
$\sin \theta \cdot \sin \theta = {\sin}^{2} \theta$ (That's true and therefore the original identity holds true)

4) ${\csc}^{2} \theta \cdot {\tan}^{2} \theta - 1 = {\tan}^{2} \theta$

$\frac{1}{\cancel{{\sin}^{2} \theta}} \cdot \frac{\cancel{{\sin}^{2} \theta}}{\cos} ^ 2 \theta - 1 = {\tan}^{2} \theta$
${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$ (We know that this is true but we can go on)
$\frac{1}{\cos} ^ 2 \theta - 1 = {\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta$
$\frac{1 - {\cos}^{2} \theta}{\cancel{{\cos}^{2} \theta}} = {\sin}^{2} \frac{\theta}{\cancel{{\cos}^{2} \theta}}$
$1 = {\sin}^{2} \theta + {\cos}^{2} \theta$ (That's true and therefore the original identity holds true)