Answer :
To find [tex]\(\sec \theta\)[/tex] given that [tex]\(\tan \theta = \frac{m^2 - n^2}{2 m n}\)[/tex], follow these steps:
1. Understand the Given Expression:
- We are given [tex]\(\tan \theta = \frac{m^2 - n^2}{2 m n}\)[/tex].
- Remember that [tex]\(\tan \theta\)[/tex] is the ratio of the opposite side to the adjacent side in a right triangle.
2. Recall the Trigonometric Identity:
- We need to use the Pythagorean identity involving [tex]\(\sec \theta\)[/tex] and [tex]\(\tan \theta\)[/tex]:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
- This can be rearranged to find [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \sqrt{1 + \tan^2 \theta} \][/tex]
3. Calculate [tex]\(\tan^2 \theta\)[/tex]:
- Since [tex]\(\tan \theta\)[/tex] is given by [tex]\(\frac{m^2 - n^2}{2 m n}\)[/tex], we calculate its square:
[tex]\[ \tan^2 \theta = \left( \frac{m^2 - n^2}{2 m n} \right)^2 \][/tex]
[tex]\[ \tan^2 \theta = \frac{(m^2 - n^2)^2}{(2 m n)^2} \][/tex]
[tex]\[ \tan^2 \theta = \frac{(m^2 - n^2)^2}{4 m^2 n^2} \][/tex]
4. Substitute [tex]\(\tan^2 \theta\)[/tex] into the Identity:
[tex]\[ \sec^2 \theta = 1 + \frac{(m^2 - n^2)^2}{4 m^2 n^2} \][/tex]
5. Simplify the Expression for [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec^2 \theta = \frac{4 m^2 n^2 + (m^2 - n^2)^2}{4 m^2 n^2} \][/tex]
[tex]\[ \sec^2 \theta = \frac{4 m^2 n^2 + m^4 - 2m^2 n^2 + n^4}{4 m^2 n^2} \][/tex]
[tex]\[ \sec^2 \theta = \frac{m^4 + 2m^2 n^2 + n^4}{4 m^2 n^2} \][/tex]
[tex]\[ \sec^2 \theta = \frac{(m^2 + n^2)^2}{4 m^2 n^2} \][/tex]
[tex]\[ \sec^2 \theta = \left( \frac{m^2 + n^2}{2 m n} \right)^2 \][/tex]
6. Take the Square Root:
- Finally, take the square root of both sides to find [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{m^2 + n^2}{2 m n} \][/tex]
From the numbers provided in the solution, [tex]\(\tan \theta\)[/tex] is approximately [tex]\(0.5333\)[/tex] and [tex]\(\sec \theta\)[/tex] is approximately [tex]\(1.1333\)[/tex].
1. Understand the Given Expression:
- We are given [tex]\(\tan \theta = \frac{m^2 - n^2}{2 m n}\)[/tex].
- Remember that [tex]\(\tan \theta\)[/tex] is the ratio of the opposite side to the adjacent side in a right triangle.
2. Recall the Trigonometric Identity:
- We need to use the Pythagorean identity involving [tex]\(\sec \theta\)[/tex] and [tex]\(\tan \theta\)[/tex]:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
- This can be rearranged to find [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \sqrt{1 + \tan^2 \theta} \][/tex]
3. Calculate [tex]\(\tan^2 \theta\)[/tex]:
- Since [tex]\(\tan \theta\)[/tex] is given by [tex]\(\frac{m^2 - n^2}{2 m n}\)[/tex], we calculate its square:
[tex]\[ \tan^2 \theta = \left( \frac{m^2 - n^2}{2 m n} \right)^2 \][/tex]
[tex]\[ \tan^2 \theta = \frac{(m^2 - n^2)^2}{(2 m n)^2} \][/tex]
[tex]\[ \tan^2 \theta = \frac{(m^2 - n^2)^2}{4 m^2 n^2} \][/tex]
4. Substitute [tex]\(\tan^2 \theta\)[/tex] into the Identity:
[tex]\[ \sec^2 \theta = 1 + \frac{(m^2 - n^2)^2}{4 m^2 n^2} \][/tex]
5. Simplify the Expression for [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec^2 \theta = \frac{4 m^2 n^2 + (m^2 - n^2)^2}{4 m^2 n^2} \][/tex]
[tex]\[ \sec^2 \theta = \frac{4 m^2 n^2 + m^4 - 2m^2 n^2 + n^4}{4 m^2 n^2} \][/tex]
[tex]\[ \sec^2 \theta = \frac{m^4 + 2m^2 n^2 + n^4}{4 m^2 n^2} \][/tex]
[tex]\[ \sec^2 \theta = \frac{(m^2 + n^2)^2}{4 m^2 n^2} \][/tex]
[tex]\[ \sec^2 \theta = \left( \frac{m^2 + n^2}{2 m n} \right)^2 \][/tex]
6. Take the Square Root:
- Finally, take the square root of both sides to find [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \frac{m^2 + n^2}{2 m n} \][/tex]
From the numbers provided in the solution, [tex]\(\tan \theta\)[/tex] is approximately [tex]\(0.5333\)[/tex] and [tex]\(\sec \theta\)[/tex] is approximately [tex]\(1.1333\)[/tex].
