if $sec(a) =frac{x}{4}$ for $0 < x < frac{pi}{2}$ , find an expression for $ln|sec(a) + tan(a)|$ in terms of $x$.

I’m given the following question and am coming up with a different answer from the book.

if $sec(a) = frac{x}{4}$ for $0 < x < frac{pi}{2}$ , find an expression for $ln|sec(a) + tan(a)|$ in terms of $x$.

The book answer is $ln|x + sqrt{x^2 + 16}| – ln(4)$ where as I’m getting $ln|x + sqrt{x^2 – 16}| – ln(4)$. What am I doing wrong?

Since $sec(a) = frac{x}{4}$ then $cos(a) = frac{4}{x}$ thus adjacent = $4$, hypotenuse = $x$ and we have to find opposite.

Because $adj^2 + opp^2 = Hyp^2$ we have $16 + opp^2$ = $x^2$. And so $opp^2 = x^2 – 16$ thus opposite = $sqrt{x^2 – 16}$

$sec(a) + tan(a) = frac{x}{4} + frac{sqrt{x^2 – 16}}{4}$ and so $ln|sec(a) + tan(a)| = ln|frac{x}{4} + frac{sqrt{x^2 – 16}}{4}| = ln|frac{x + sqrt{x^2 – 16}}{4}| = ln|x + sqrt{x^2 – 16}| – ln(4)$