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

Mathematics Asked by maybedave on January 2, 2021

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