Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Certainly! Let's solve the equation step-by-step:
Given the equation:
[tex]\[ 7 \sec^2 \theta + \tan^2 \theta = 15 \][/tex]
First, recall that the secant function and the tangent function have a trigonometric identity relationship. Specifically, for any angle [tex]\(\theta\)[/tex]:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
We will use this identity to simplify our given equation. Substitute [tex]\(\sec^2 \theta\)[/tex] with [tex]\(1 + \tan^2 \theta\)[/tex]:
[tex]\[ 7 (1 + \tan^2 \theta) + \tan^2 \theta = 15 \][/tex]
Expand and combine like terms:
[tex]\[ 7 + 7 \tan^2 \theta + \tan^2 \theta = 15 \][/tex]
[tex]\[ 7 + 8 \tan^2 \theta = 15 \][/tex]
Isolate the term involving [tex]\(\tan^2 \theta\)[/tex]:
[tex]\[ 8 \tan^2 \theta = 15 - 7 \][/tex]
[tex]\[ 8 \tan^2 \theta = 8 \][/tex]
Now divide both sides by 8:
[tex]\[ \tan^2 \theta = 1 \][/tex]
Taking the square root of both sides, we get:
[tex]\[ \tan \theta = \pm 1 \][/tex]
We now need to determine the angles [tex]\(\theta\)[/tex] where the tangent is [tex]\(\pm 1\)[/tex]. Recall the values of [tex]\(\theta\)[/tex] in one full circle (from [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex] radians) where this occurs:
1. [tex]\(\tan \theta = 1\)[/tex] at [tex]\(\theta = \frac{\pi}{4} \)[/tex] and [tex]\(\theta = \frac{5\pi}{4}\)[/tex].
2. [tex]\(\tan \theta = -1\)[/tex] at [tex]\(\theta = \frac{3\pi}{4}\)[/tex] and [tex]\(\theta = \frac{7\pi}{4}\)[/tex].
However, [tex]\(\tan \theta\)[/tex] repeats every [tex]\(\pi\)[/tex] radians. Therefore, we can simply use the general solutions for these values within the range of [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex]. Thus, the angles are:
[tex]\[ \theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4} \][/tex]
To also consider angles in other quadrants, let's include the negative rotations:
[tex]\[ \theta = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4} \][/tex]
Therefore, the complete set of solutions is:
[tex]\[ \theta = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4} \][/tex]
These are the angles [tex]\(\theta\)[/tex] where the given equation [tex]\(7 \sec^2 \theta + \tan^2 \theta = 15\)[/tex] holds true.
Given the equation:
[tex]\[ 7 \sec^2 \theta + \tan^2 \theta = 15 \][/tex]
First, recall that the secant function and the tangent function have a trigonometric identity relationship. Specifically, for any angle [tex]\(\theta\)[/tex]:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
We will use this identity to simplify our given equation. Substitute [tex]\(\sec^2 \theta\)[/tex] with [tex]\(1 + \tan^2 \theta\)[/tex]:
[tex]\[ 7 (1 + \tan^2 \theta) + \tan^2 \theta = 15 \][/tex]
Expand and combine like terms:
[tex]\[ 7 + 7 \tan^2 \theta + \tan^2 \theta = 15 \][/tex]
[tex]\[ 7 + 8 \tan^2 \theta = 15 \][/tex]
Isolate the term involving [tex]\(\tan^2 \theta\)[/tex]:
[tex]\[ 8 \tan^2 \theta = 15 - 7 \][/tex]
[tex]\[ 8 \tan^2 \theta = 8 \][/tex]
Now divide both sides by 8:
[tex]\[ \tan^2 \theta = 1 \][/tex]
Taking the square root of both sides, we get:
[tex]\[ \tan \theta = \pm 1 \][/tex]
We now need to determine the angles [tex]\(\theta\)[/tex] where the tangent is [tex]\(\pm 1\)[/tex]. Recall the values of [tex]\(\theta\)[/tex] in one full circle (from [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex] radians) where this occurs:
1. [tex]\(\tan \theta = 1\)[/tex] at [tex]\(\theta = \frac{\pi}{4} \)[/tex] and [tex]\(\theta = \frac{5\pi}{4}\)[/tex].
2. [tex]\(\tan \theta = -1\)[/tex] at [tex]\(\theta = \frac{3\pi}{4}\)[/tex] and [tex]\(\theta = \frac{7\pi}{4}\)[/tex].
However, [tex]\(\tan \theta\)[/tex] repeats every [tex]\(\pi\)[/tex] radians. Therefore, we can simply use the general solutions for these values within the range of [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex]. Thus, the angles are:
[tex]\[ \theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4} \][/tex]
To also consider angles in other quadrants, let's include the negative rotations:
[tex]\[ \theta = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4} \][/tex]
Therefore, the complete set of solutions is:
[tex]\[ \theta = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4} \][/tex]
These are the angles [tex]\(\theta\)[/tex] where the given equation [tex]\(7 \sec^2 \theta + \tan^2 \theta = 15\)[/tex] holds true.
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.