Question #0042b

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Question #0042b

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Answer 1 · 2017-08-24T22:22:43

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

Explanation:

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

For all other $θ$ $theta$ , we have

$(1 - {sin}^{2} θ) \cdot \frac{1}{{cos}^{2} θ} = {cos}^{2} θ \cdot \frac{1}{{cos}^{2} θ} = 1$ $(1-sin^2 theta) * 1/cos^2 theta = cos^2 theta * 1/cos^2 theta = 1$

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

Answer 2 · 2017-08-24T22:22:53

See explanation below.

Explanation:

We have: $(1 - {sin}^{2} (θ)) {sec}^{2} (θ)$ $(1 - sin^(2)(theta)) sec^(2)(theta)$

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

We can rearrange it to get:

$\Rightarrow {cos}^{2} (θ) = 1 - {sin}^{2} (θ)$ $Rightarrow cos^(2)(theta) = 1 - sin^(2)(theta)$

Let's apply this rearranged identity to our proof:

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

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

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

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

$= 1$ $= 1 " " "$ $Q.E.D.$ $"Q.E.D."$

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Question #0042b

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