Understanding the Problem: Finding sec θ

The question provides a relationship between the secant and tangent of an angle $\theta$: $\sec \theta + \tan \theta = 5$. We are asked to find the value of $\sec \theta$ given this information. The constraint $\theta \ne 0$ is mentioned, which is important in some contexts but for this specific calculation using the fundamental identity, it doesn't change the approach significantly.

Using Trigonometric Identities to Solve for sec θ

A key trigonometric identity relates secant and tangent: $\sec^2 \theta - \tan^2 \theta = 1$. This identity holds true for all angles where $\sec \theta$ and $\tan \theta$ are defined.

We can factor the left side of this identity as a difference of squares:

$\sec^2 \theta - \tan^2 \theta = (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)$

So, the identity becomes:

$(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$

Applying the Given Information

We are given that $\sec \theta + \tan \theta = 5$. We can substitute this value into the factored identity:

$(\sec \theta - \tan \theta)(5) = 1$

Solving for sec θ - tan θ

Now we can easily find the value of $\sec \theta - \tan \theta$ by dividing both sides by 5:

$\sec \theta - \tan \theta = \frac{1}{5}$

Solving the System of Equations for sec θ

We now have two linear equations involving $\sec \theta$ and $\tan \theta$:

1. Equation 1: $\sec \theta + \tan \theta = 5$
2. Equation 2: $\sec \theta - \tan \theta = \frac{1}{5}$

To find $\sec \theta$, we can add these two equations. Notice that the $\tan \theta$ terms will cancel out:

$(\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = 5 + \frac{1}{5}$

Combining like terms on the left side:

$2 \sec \theta = 5 + \frac{1}{5}$

To find $\sec \theta$, we divide both sides by 2:

$\sec \theta = \frac{1}{2} \left( 5 + \frac{1}{5} \right)$

Comparing with Options

Let's compare our result with the given options:

• Option 1: $(3 + \frac{1}{3})$
• Option 2: $\frac{1}{2}(5 + \frac{1}{5})$
• Option 3: $\frac{1}{2}(3 + \frac{1}{3})$
• Option 4: $(5 + \frac{1}{5})$

Our calculated value for $\sec \theta$ matches Option 2.

Step-by-Step Calculation Summary

Here's a summary of the steps:

1. Start with the given equation: $\sec \theta + \tan \theta = 5$.
2. Use the identity: $\sec^2 \theta - \tan^2 \theta = 1$.
3. Factor the identity: $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
4. Substitute the given value: $(\sec \theta - \tan \theta)(5) = 1$.
5. Solve for the difference: $\sec \theta - \tan \theta = \frac{1}{5}$.
6. Add the original equation and the difference equation: $(\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = 5 + \frac{1}{5}$.
7. Simplify and solve for $2 \sec \theta$: $2 \sec \theta = 5 + \frac{1}{5}$.
8. Solve for $\sec \theta$: $\sec \theta = \frac{1}{2} \left( 5 + \frac{1}{5} \right)$.

Revision Table: Trigonometry Identities

Reviewing fundamental identities is crucial for solving trigonometry problems.

Additional Information: Alternative Methods

While the method using the identity $\sec^2 \theta - \tan^2 \theta = 1$ is the most common for this type of problem, one could potentially solve it by converting everything to sine and cosine, although it would likely be more complex.

Given $\sec \theta + \tan \theta = 5$, we can write:

$\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = 5$

$\frac{1 + \sin \theta}{\cos \theta} = 5$

$1 + \sin \theta = 5 \cos \theta$

Squaring both sides:

$(1 + \sin \theta)^2 = (5 \cos \theta)^2$

$1 + 2 \sin \theta + \sin^2 \theta = 25 \cos^2 \theta$

Using $\cos^2 \theta = 1 - \sin^2 \theta$:

$1 + 2 \sin \theta + \sin^2 \theta = 25 (1 - \sin^2 \theta)$

$1 + 2 \sin \theta + \sin^2 \theta = 25 - 25 \sin^2 \theta$

Rearranging into a quadratic equation in terms of $\sin \theta$:

$26 \sin^2 \theta + 2 \sin \theta - 24 = 0$

Divide by 2:

$13 \sin^2 \theta + \sin \theta - 12 = 0$

This quadratic equation can be solved for $\sin \theta$. Using the quadratic formula $\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=13$, $b=1$, $c=-12$:

$\sin \theta = \frac{-1 \pm \sqrt{1^2 - 4(13)(-12)}}{2(13)}$

$\sin \theta = \frac{-1 \pm \sqrt{1 + 624}}{26}$

$\sin \theta = \frac{-1 \pm \sqrt{625}}{26}$

$\sin \theta = \frac{-1 \pm 25}{26}$

This gives two possible values for $\sin \theta$:

• $\sin \theta = \frac{-1 + 25}{26} = \frac{24}{26} = \frac{12}{13}$
• $\sin \theta = \frac{-1 - 25}{26} = \frac{-26}{26} = -1$

If $\sin \theta = -1$, then $\theta = 270^\circ$ (or $3\pi/2$ radians), which means $\cos \theta = 0$. However, $\sec \theta = 1/\cos \theta$ and $\tan \theta = \sin \theta/\cos \theta$ would be undefined. So $\sin \theta = -1$ is not a valid solution in this context.

Using $\sin \theta = \frac{12}{13}$, we can find $\cos \theta$ using $\cos^2 \theta = 1 - \sin^2 \theta$:

$\cos^2 \theta = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$

$\cos \theta = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13}$

Now we check which value of $\cos \theta$ satisfies the original equation $\frac{1 + \sin \theta}{\cos \theta} = 5$ with $\sin \theta = 12/13$:

• If $\cos \theta = \frac{5}{13}$: $\frac{1 + 12/13}{5/13} = \frac{(13+12)/13}{5/13} = \frac{25/13}{5/13} = \frac{25}{5} = 5$. This works.
• If $\cos \theta = -\frac{5}{13}$: $\frac{1 + 12/13}{-5/13} = \frac{25/13}{-5/13} = \frac{25}{-5} = -5$. This does not work.

So, $\sin \theta = \frac{12}{13}$ and $\cos \theta = \frac{5}{13}$.

Finally, $\sec \theta = \frac{1}{\cos \theta} = \frac{1}{5/13} = \frac{13}{5}$.

Let's verify if $\frac{13}{5}$ is equal to $\frac{1}{2} \left( 5 + \frac{1}{5} \right)$:

$\frac{1}{2} \left( 5 + \frac{1}{5} \right) = \frac{1}{2} \left( \frac{25}{5} + \frac{1}{5} \right) = \frac{1}{2} \left( \frac{26}{5} \right) = \frac{26}{10} = \frac{13}{5}$.

Both methods yield the same result for $\sec \theta$. The first method using the identity $\sec^2 \theta - \tan^2 \theta = 1$ is much quicker and more direct for this specific problem format.