This is Exercise 1 in I 3.5 asking you to prove Theorem I.24, and you are permitted to use earlier theorems and axioms.
Firstly, your argument to prove that $a$ and $b$ are nonzero is incorrect. This follows directly from one of the earlier theorems, but not from Theorem I.11 as you say. Can you identify the right one?
Next, assume for a contradiction that $a$ and $b$ are not both positive or both negative, that is that they have opposite signs. For example, since $a$ and $b$ play similar roles, we may assume that $a > 0$ and $b < 0$. (When I say that $a$ and $b$ play similar roles, I am relying on the fact that $ab = ba$.)
Now you can apply Theorem I.19 to the inequality $b < 0$, multiplying by $a$, which is positive. (We again need the same earlier theorem here.) This will yield a contradiction, proving what you want.
The reason a proof by contradiction is best, in my opinion, is that it avoids introducing consideration of inverses. In other words, you make certain assumptions about the signs of $a$ and $b$, and you determine the sign of $ab$. What really matters here are the cases where $ab$ is negative. Since your assumption is $ab > 0$, this means you must argue by contradiction.
More precisely, you are proving the contrapositive of the original statement, but this is a technicality.
$\sqrt{\frac{a+x}{a-x}}$, $a>0$, implies $\frac{a+x}{a-x} \geq 0$
So we have, $\frac{a+x}{a-x}=0$, implying $a+x=0$, which is trivial and uninteresting since $\int 0 dx=C$. Meaning, this is a special case for some particular value of $x$.
Or, we have $b>\frac{a+x}{a-x}>0$ for some $b\in \mathbb R^{+}$ which implies $b(a-x)>(a+x)>0$ which implies $a>-x$
This is not limited to end-points, we can generalize it to an arbitraty discontinuity that is isolated, given one trivial assumption that $F(x)$ is defined for $x=-a$, consider the second FTC:
Given a continous, integrable, function $f(x)$ on some interval $[a,b]$, $\int_a^bf(x)=F(b)-F(a)$, given that $F'(x)=f(x)$.
In this case, suppose our function g(x) is continous, and integrable, on $[a,b]\setminus \{c\}$ which means $g(x)$ is integrable on $[a, c- \epsilon ] \cup [c + \epsilon ,b]$.
We get, $$\lim \limits_{d^{-} \to c}\int_a^dg(x)dx + \lim \limits_{e^{+} \to c}\int_e^b g(x)dx = \lim \limits_{d^{-}\to c}G(d)-G(a)+G(b)-\lim \limits_{e^{+}\to c}G(e)=G(c)-G(a)+G(b)-G(c)=G(b)-G(a)$$
The assumption here is that the two limits exist and are equal, which is guaranteed by the fact that $a=-x$ is a special case for some particular value of $x$. If you integrate the the function, $\sqrt{\frac{a+x}{a-x}}$, under the assumption that $x\neq -a$ and check the Limit, $x\to -a$ you can see that it exists, and is one and the same from both directions, then you can discharge the assumption.
Best Answer
We have
$$f(n)-f(n+1)=\int_{0}^{\frac{\pi}{4}}\left(\tan^n(x)-\tan^{n+1}(x)\right)dx$$ $$=\int_{0}^{\frac{\pi}{4}}\tan^{n}(x)\left(1-\tan(x)\right)dx>0$$
since $0<\tan(x)<1$ for $0<x<\frac{\pi}{4},$ so the integral is positive because the integrand is positive on the interval of integration. Hence $f(n)>f(n+1)$ as required.