Question #0042b

Aug 24, 2017

The solution is: all $\theta$ except of $\theta = \frac{\pi}{2} + \pi k$ with $k$ an integer.

Explanation:

${\sec}^{2} \theta = \frac{1}{\cos} ^ 2 \theta$ is not defined for $\cos \theta = 0$, so it is not defined for $\theta = \frac{\pi}{4} + \pi k$ with $k$ an integer.

For all other $\theta$, we have

$\left(1 - {\sin}^{2} \theta\right) \cdot \frac{1}{\cos} ^ 2 \theta = {\cos}^{2} \theta \cdot \frac{1}{\cos} ^ 2 \theta = 1$

So, every other $\theta$ is a solution.

Aug 24, 2017

See explanation below.

Explanation:

We have: $\left(1 - {\sin}^{2} \left(\theta\right)\right) {\sec}^{2} \left(\theta\right)$

One of the Pythagorean identities is ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$.

We can rearrange it to get:

$R i g h t a r r o w {\cos}^{2} \left(\theta\right) = 1 - {\sin}^{2} \left(\theta\right)$

Let's apply this rearranged identity to our proof:

$= {\cos}^{2} \left(\theta\right) \cdot {\sec}^{2} \left(\theta\right)$

Let's apply the standard trigonometric identity $\sec \left(\theta\right) = \frac{1}{\cos \left(\theta\right)}$:

$= {\cos}^{2} \left(\theta\right) \cdot \frac{1}{{\cos}^{2} \left(\theta\right)}$

$= \frac{{\cos}^{2} \left(\theta\right)}{{\cos}^{2} \left(\theta\right)}$

$= 1 \text{ " }$ $\text{Q.E.D.}$