Sure, let's solve the equation step-by-step to prove that [tex]\(\sec \theta = \pm \frac{\sqrt{17}}{4}\)[/tex].
We start with the equation:
[tex]\[
\frac{\cos \theta}{\tan \theta \cdot \sin \theta} = 16
\][/tex]
First, we'll simplify the left-hand side by expressing [tex]\(\tan \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
Now substitute [tex]\(\tan \theta\)[/tex] back into the original equation:
[tex]\[
\frac{\cos \theta}{\left(\frac{\sin \theta}{\cos \theta}\right) \cdot \sin \theta} = 16
\][/tex]
Simplify the denominator:
[tex]\[
\frac{\cos \theta}{\frac{\sin \theta^2}{\cos \theta}} = 16
\][/tex]
When we divide by a fraction, we multiply by its reciprocal:
[tex]\[
\frac{\cos \theta \cdot \cos \theta}{\sin \theta^2} = 16
\][/tex]
[tex]\[
\frac{\cos^2 \theta}{\sin^2 \theta} = 16
\][/tex]
This can be expressed as:
[tex]\[
\cot^2 \theta = 16
\][/tex]
Now, solving for [tex]\(\cot \theta\)[/tex]:
[tex]\[
\cot \theta = \pm 4
\][/tex]
We know that:
[tex]\[
\cot \theta = \frac{\cos \theta}{\sin \theta}
\][/tex]
So:
[tex]\[
\frac{\cos \theta}{\sin \theta} = \pm 4
\][/tex]
Cross-multiplying:
[tex]\[
\cos \theta = \pm 4 \sin \theta
\][/tex]
Next, let's find an expression for [tex]\(\sec \theta\)[/tex]. Recall that [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex]. Using the Pythagorean identity, [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
First, let's express [tex]\(\sin \theta\)[/tex] in terms of [tex]\(\cos \theta\)[/tex]:
[tex]\[
\sin^2 \theta = 1 - \cos^2 \theta
\][/tex]
Substitute [tex]\(\cos \theta = \pm 4 \sin \theta\)[/tex] into [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[
\sin^2 \theta + (4 \sin \theta)^2 = 1
\][/tex]
[tex]\[
\sin^2 \theta + 16 \sin^2 \theta = 1
\][/tex]
[tex]\[
17 \sin^2 \theta = 1
\][/tex]
[tex]\[
\sin^2 \theta = \frac{1}{17}
\][/tex]
Thus:
[tex]\[
\sin \theta = \pm \frac{1}{\sqrt{17}}
\][/tex]
Now substitute [tex]\(\sin \theta = \pm \frac{1}{\sqrt{17}}\)[/tex] back into [tex]\(\cos \theta = \pm 4 \sin \theta\)[/tex]:
[tex]\[
\cos \theta = \pm 4 \left(\pm \frac{1}{\sqrt{17}}\right)
\][/tex]
[tex]\[
\cos \theta = \pm \frac{4}{\sqrt{17}}
\][/tex]
Finally, we find [tex]\(\sec \theta\)[/tex]:
[tex]\[
\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\pm \frac{4}{\sqrt{17}}} = \pm \frac{\sqrt{17}}{4}
\][/tex]
Hence, we have proven that:
[tex]\[
\sec \theta = \pm \frac{\sqrt{17}}{4}
\][/tex]
