If tan θ + sec θ = 3, then what is the value of 3 tan θ + 9 sec θ?
19
Solving Trigonometric Equations with tan \( \theta \) and sec \( \theta \)
The problem provides us with a relationship between \( \tan \theta \) and \( \sec \theta \) and asks us to find the value of a linear expression involving these functions.
We are given the equation:
\( \tan \theta + \sec \theta = 3 \)
We need to find the value of \( 3 \tan \theta + 9 \sec \theta \).
To find the value of this expression, we first need to determine the individual values of \( \tan \theta \) and \( \sec \theta \).
Using Trigonometric Identities
A fundamental trigonometric identity relates \( \tan \theta \) and \( \sec \theta \):
\( \sec^2 \theta - \tan^2 \theta = 1 \)
This identity can be factored as a difference of squares:
\( (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 \)
We are given that \( \tan \theta + \sec \theta = 3 \). We can substitute this into the factored identity:
\( (\sec \theta - \tan \theta)(3) = 1 \)
Now, we can solve for \( (\sec \theta - \tan \theta) \):
\( \sec \theta - \tan \theta = \frac{1}{3} \)
Solving for tan \( \theta \) and sec \( \theta \)
We now have a system of two linear equations involving \( \tan \theta \) and \( \sec \theta \):
- \( \sec \theta + \tan \theta = 3 \)
- \( \sec \theta - \tan \theta = \frac{1}{3} \)
We can solve this system by adding and subtracting the two equations.
Adding the equations:
\( (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = 3 + \frac{1}{3} \)
\( 2 \sec \theta = \frac{9}{3} + \frac{1}{3} \)
\( 2 \sec \theta = \frac{10}{3} \)
\( \sec \theta = \frac{10}{3 \times 2} = \frac{10}{6} = \frac{5}{3} \)
Subtracting the second equation from the first:
\( (\sec \theta + \tan \theta) - (\sec \theta - \tan \theta) = 3 - \frac{1}{3} \)
\( \sec \theta + \tan \theta - \sec \theta + \tan \theta = \frac{9}{3} - \frac{1}{3} \)
\( 2 \tan \theta = \frac{8}{3} \)
\( \tan \theta = \frac{8}{3 \times 2} = \frac{8}{6} = \frac{4}{3} \)
So, we have found the values:
- \( \tan \theta = \frac{4}{3} \)
- \( \sec \theta = \frac{5}{3} \)
Calculating the Final Expression Value
Now we need to find the value of \( 3 \tan \theta + 9 \sec \theta \). Substitute the values we found:
\( 3 \tan \theta + 9 \sec \theta = 3 \times \left(\frac{4}{3}\right) + 9 \times \left(\frac{5}{3}\right) \)
Simplify the expression:
\( = \left(3 \times \frac{4}{3}\right) + \left(9 \times \frac{5}{3}\right) \)
\( = 4 + \left(\frac{9}{3} \times 5\right) \)
\( = 4 + (3 \times 5) \)
\( = 4 + 15 \)
\( = 19 \)
The value of \( 3 \tan \theta + 9 \sec \theta \) is 19.
Summary of Steps
Here is a quick summary of the steps taken to solve this problem:
- Used the given equation \( \tan \theta + \sec \theta = 3 \).
- Used the identity \( \sec^2 \theta - \tan^2 \theta = 1 \), factored it into \( (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 \).
- Substituted the given value into the identity to find \( \sec \theta - \tan \theta = \frac{1}{3} \).
- Solved the system of equations \( \sec \theta + \tan \theta = 3 \) and \( \sec \theta - \tan \theta = \frac{1}{3} \) to find \( \tan \theta = \frac{4}{3} \) and \( \sec \theta = \frac{5}{3} \).
- Substituted these values into the expression \( 3 \tan \theta + 9 \sec \theta \) and calculated the final value, which is 19.
Revision Table: Key Trigonometric Concepts
| Concept | Description | Identity |
|---|---|---|
| Tangent (tan \( \theta \)) | Ratio of opposite side to adjacent side in a right triangle. Also \( \frac{\sin \theta}{\cos \theta} \). | |
| Secant (sec \( \theta \)) | Ratio of hypotenuse to adjacent side in a right triangle. Reciprocal of cosine: \( \frac{1}{\cos \theta} \). | |
| Pythagorean Identity | Relates secant and tangent squared. | \( \sec^2 \theta - \tan^2 \theta = 1 \) |
Additional Information: Relationship Between tan \( \theta \) and sec \( \theta \)
The identity \( \sec^2 \theta - \tan^2 \theta = 1 \) is derived directly from the primary Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). If you divide the primary identity by \( \cos^2 \theta \) (assuming \( \cos \theta \neq 0 \)), you get:
\( \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} \)
This simplifies to:
\( \left(\frac{\sin \theta}{\cos \theta}\right)^2 + 1 = \left(\frac{1}{\cos \theta}\right)^2 \)
\( \tan^2 \theta + 1 = \sec^2 \theta \)
Rearranging this gives the identity used in the solution:
\( \sec^2 \theta - \tan^2 \theta = 1 \)
This identity is valid for all angles \( \theta \) where \( \cos \theta \neq 0 \). Problems like the one solved here often rely on using this fundamental relationship and algebraic manipulation, such as factoring the difference of squares.
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