All $$x$$ satisfying the inequality $$(\cot^{-1} x)^2 - 7(\cot^{-1} x) + 10 > 0$$, lie in the interval:

We begin by introducing a simpler symbol. Put

$$y=\cot^{-1}x.$$

By definition of the principal value, the inverse cotangent always satisfies

$$0<y<\pi.$$

Substituting $$y$$ into the given inequality we obtain

$$y^{2}-7y+10>0.$$

Now we treat this as an ordinary quadratic inequality in $$y$$. First we factor the left-hand side:

$$y^{2}-7y+10=(y-5)(y-2).$$

The general rule for a product of two real factors is:

• If $$(y-a)(y-b)>0$$, then $$y<\min\{a,b\}\quad\text{or}\quad y>\max\{a,b\}.$$

Applying the rule with $$a=5,\;b=2$$ gives

$$y<2\quad\text{or}\quad y>5.$$

But we must keep in mind the principal-value restriction $$0<y<\pi\;(=3.14159\ldots).$$ The part $$y>5$$ is impossible because $$5>\pi$$, so it is thrown away. The only admissible portion is

$$0<y<2.$$

So far we have shown that

$$0<\cot^{-1}x<2.$$

We now convert this back to a statement about $$x$$. For that we recall a key fact about the inverse cotangent:

• The function $$y=\cot^{-1}x$$ is strictly decreasing on the entire real line. In symbols, if $$x_{1}<x_{2},$$ then $$\cot^{-1}x_{1}>\cot^{-1}x_{2}.$$

Because of this monotonic behaviour, reversing an inequality inside $$\cot^{-1}$$ flips the inequality sign. From

$$\cot^{-1}x<2$$

we therefore get

$$x>\cot 2.$$

(The value $$\cot 2$$ is a fixed negative number, approximately $$-0.4577$$.)

No further restriction is necessary, since every real $$x$$ already forces $$\cot^{-1}x>0$$ automatically. Consequently the complete solution set in terms of $$x$$ is

$$\boxed{(\,\cot 2,\;\infty\,)}.$$

When we compare this with the four alternatives supplied in the question, we see that it matches Option B.

Hence, the correct answer is Option 2.