Sure, let's solve the problem step-by-step:
Given:
[tex]\[ \tan \theta + \sec \theta = P \][/tex]
We need to find an expression for [tex]\(\sec \theta\)[/tex] in terms of [tex]\(P\)[/tex].
1. Notation and Assumptions:
Let's denote [tex]\(\sec \theta\)[/tex] by [tex]\(y\)[/tex]. Then:
[tex]\[ \sec \theta = y \][/tex]
2. Express [tex]\(\tan \theta\)[/tex] in Terms of [tex]\(y\)[/tex]:
Recall the Pythagorean identity involving [tex]\(\sec \theta\)[/tex] and [tex]\(\tan \theta\)[/tex]:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]
Since [tex]\(\sec \theta = y\)[/tex], we have:
[tex]\[ y^2 = 1 + \tan^2 \theta \][/tex]
Solving for [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan^2 \theta = y^2 - 1 \][/tex]
[tex]\[ \tan \theta = \sqrt{y^2 - 1} \][/tex]
3. Form the Given Equation:
Substitute [tex]\(\tan \theta\)[/tex] and [tex]\(\sec \theta\)[/tex] in the given equation:
[tex]\[ \sqrt{y^2 - 1} + y = P \][/tex]
4. Solve for [tex]\(y\)[/tex] (Secant Theta):
Isolate the square root term:
[tex]\[ \sqrt{y^2 - 1} = P - y \][/tex]
Square both sides to eliminate the square root:
[tex]\[ y^2 - 1 = (P - y)^2 \][/tex]
Expand the right-hand side:
[tex]\[ y^2 - 1 = P^2 - 2Py + y^2 \][/tex]
Subtract [tex]\(y^2\)[/tex] from both sides:
[tex]\[ -1 = P^2 - 2Py \][/tex]
Rearrange the equation to solve for [tex]\(y\)[/tex]:
[tex]\[ 2Py = P^2 + 1 \][/tex]
[tex]\[ y = \frac{P^2 + 1}{2P} \][/tex]
Thus, we have:
[tex]\[ \sec \theta = \frac{P^2 + 1}{2P} \][/tex]
So, the expression for [tex]\(\sec \theta\)[/tex] in terms of [tex]\(P\)[/tex] is:
[tex]\[ \sec \theta = \frac{P^2 + 1}{2P} \][/tex]
