Sure, let's work through solving the trigonometric equation [tex]\((\cos \theta) \cot \theta + \sin \theta = \sec \theta\)[/tex].
First, recall the definitions in terms of sine and cosine:
- [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex]
- [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex]
Substitute these identities into the equation:
[tex]\[
(\cos \theta) \left(\frac{\cos \theta}{\sin \theta}\right) + \sin \theta = \frac{1}{\cos \theta}
\][/tex]
This simplifies to:
[tex]\[
\frac{\cos^2 \theta}{\sin \theta} + \sin \theta = \frac{1}{\cos \theta}
\][/tex]
We need a common denominator to combine the terms on the left-hand side. The common denominator will be [tex]\(\sin \theta\)[/tex]:
[tex]\[
\frac{\cos^2 \theta + \sin^2 \theta \cdot \sin \theta}{\sin \theta} = \frac{1}{\cos \theta}
\][/tex]
Recognize that [tex]\(\cos^2 \theta + \sin^2 \theta = 1\)[/tex] by the Pythagorean identity, so:
[tex]\[
\frac{1}{\sin \theta} = \frac{1}{\cos \theta}
\][/tex]
Set the equivalent numerators equal:
[tex]\[
\sin \theta = \cos \theta
\][/tex]
We solve for [tex]\(\theta\)[/tex] where [tex]\(\sin \theta = \cos \theta\)[/tex]. This occurs when [tex]\(\theta\)[/tex] results in an angle where their sines and cosines are equal. One known solution occurs at:
[tex]\[
\theta = \frac{\pi}{4}
\][/tex]
However, we should consider the periodic nature of trigonometric functions. For cosine and sine to be equal, the angles can also differ by [tex]\(\pi\)[/tex]. Thus, other solutions are in the form:
[tex]\[
\theta = \frac{\pi}{4} + k\pi \quad \text{where } k \text{ is any integer}
\][/tex]
We are particularly interested in specific solutions within the main cycle of [tex]\([0, 2\pi)\)[/tex]. By examining [tex]\(k\pi\)[/tex] shifts:
[tex]\[
\theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{\pi}{4} + \pi = \frac{\pi}{4} + \frac{4\pi}{4} = \frac{5\pi}{4}
\][/tex]
Note that angles can be expressed negatively as well through cycle shifting, producing:
[tex]\[
\theta = -\frac{3\pi}{4} \quad (\text{another form})
\][/tex]
Therefore, the solutions within one cycle [tex]\([0, 2\pi)\)[/tex] for the given equation are:
[tex]\[
\theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{5\pi}{4}
\][/tex]
But including the negative form, we get:
[tex]\[
\theta = \frac{\pi}{4} \quad \text{and} \quad \theta = -\frac{3\pi}{4}
\][/tex]
So the final solutions are:
[tex]\[
\theta = \frac{\pi}{4} \quad \text{and} \quad \theta = -\frac{3\pi}{4}
\][/tex]
