Track the order of an approximation through division

Both copies give the full chain and callouts:

$\begin{aligned}&\frac{\sin\theta}{\cos2\theta}\approx\frac{\theta}{1-(2\theta)^2/2}\\ & =\frac{\theta}{1-2\theta^2}\\ & =\theta(1-2\theta^2)^{-1}\end{aligned}$

$=\theta[1+(-1)(-2\theta^2)+\cdots]\approx\theta.$

The source recommends the standard binomial formula after bringing the denominator up as a negative power. Since the outer factor is θ, the first correction is cubic and lies beyond the retained order. Its callout’s reference to x² is read as the requested quadratic order in θ.

Both copies print

$\begin{aligned}&\frac{\theta\tan\theta}{1-\cos\theta}\approx\frac{\theta^2}{1-(1-\theta^2/2)}\\ & =\frac{\theta^2}{\theta^2/2}\\ & =2.\end{aligned}$

This preserves the source’s answer 2 as a leading-order approximation for small nonzero θ. The original quotient is undefined at θ=0; its limit there is 2.

If the instruction is read as asking for every quadratic term of the final quotient, the printed chain is insufficient. Division by an expression vanishing quadratically lowers the available order, so fourth-degree input terms are needed:

$\theta\tan\theta=\theta^2+\frac13\theta^4+O(\theta^6),\qquad 1-\cos\theta=\frac12\theta^2-\frac1{24}\theta^4+O(\theta^6).$

For small nonzero θ, cancel the common quadratic factor and expand:

$\begin{aligned}&\frac{\theta\tan\theta}{1-\cos\theta}\\ & =2\frac{1+\theta^2/3+O(\theta^4)}{1-\theta^2/12+O(\theta^4)}\\ & =2+\frac56\theta^2+O(\theta^4).\end{aligned}$

This independent precision clarification does not replace the preserved printed leading answer. Truncating the individual functions at quadratic order before a vanishing-denominator division can discard a required quadratic term of the final result.