Finding tan α from sec α + tan α

We are given the equation:

\(\sec \alpha + \tan \alpha = p \quad \text{(Equation 1)}\)

We need to find the value of \(\tan \alpha\).

We know a fundamental trigonometric identity that relates \(\sec \alpha\) and \(\tan \alpha\):

\(\sec^2 \alpha - \tan^2 \alpha = 1\)

This identity is in the form of a difference of squares, \(a^2 - b^2 = (a-b)(a+b)\). Applying this, we get:

\((\sec \alpha - \tan \alpha)(\sec \alpha + \tan \alpha) = 1\)

Now, substitute the given information from Equation 1 into this identity:

\((\sec \alpha - \tan \alpha)(p) = 1\)

From this, we can find an expression for \(\sec \alpha - \tan \alpha\):

\(\sec \alpha - \tan \alpha = \frac{1}{p} \quad \text{(Equation 2)}\)

Now we have a system of two linear equations involving \(\sec \alpha\) and \(\tan \alpha\):

• Equation 1: \(\sec \alpha + \tan \alpha = p\)
• Equation 2: \(\sec \alpha - \tan \alpha = \frac{1}{p}\)

To find \(\tan \alpha\), we can subtract Equation 2 from Equation 1. This will eliminate \(\sec \alpha\):

\((\sec \alpha + \tan \alpha) - (\sec \alpha - \tan \alpha) = p - \frac{1}{p}\)

Simplify the left side:

\(\sec \alpha + \tan \alpha - \sec \alpha + \tan \alpha = 2 \tan \alpha\)

Simplify the right side by finding a common denominator:

\(p - \frac{1}{p} = \frac{p \cdot p}{p} - \frac{1}{p} = \frac{p^2}{p} - \frac{1}{p} = \frac{p^2 - 1}{p}\)

So, the equation becomes:

\(2 \tan \alpha = \frac{p^2 - 1}{p}\)

Finally, divide by 2 to solve for \(\tan \alpha\):

\(\tan \alpha = \frac{p^2 - 1}{2p}\)

This is the value of \(\tan \alpha\) in terms of \(p\).