Question

For an acute angle $\theta$, if $\cos \theta = \frac{1}{8}$, then $\frac{8\sec\theta + 1}{8\sec\theta - 1}$ equals

• A. $\frac{64}{63}$
• B. $0$
• C. $\frac{65}{63}$
• D. $1$

Hint:

Converting secant to cosine inside the expression can also simplify the calculation directly:
\[ \frac{\frac{8}{\cos\theta} + 1}{\frac{8}{\cos\theta} - 1} = \frac{8 + \cos\theta}{8 - \cos\theta} \]
Substituting $\cos\theta = \frac{1}{8}$ gives $\frac{8 + 1/8}{8 - 1/8} = \frac{65/8}{63/8} = \frac{65}{63}$.

Answer 1

The Correct Option is C

Step 1: Rewrite the expression in terms of cos theta only.
Since $\sec\theta = \frac{1}{\cos\theta}$, multiply numerator and denominator of $\frac{8\sec\theta+1}{8\sec\theta-1}$ by $\cos\theta$ to get $\frac{8 + \cos\theta}{8 - \cos\theta}$, avoiding the need to compute $\sec\theta$ as a separate number first.
Step 2: Substitute the given value of cos theta.
We are given $\cos\theta = \frac{1}{8}$, so the expression becomes $\frac{8 + \frac{1}{8}}{8 - \frac{1}{8}}$.
Step 3: Simplify the compound fraction.
Writing both parts with denominator 8, this is $\frac{\frac{65}{8}}{\frac{63}{8}} = \frac{65}{63}$.
\[ \boxed{\frac{65}{63}} \]