<s>[INST] Trigonometric near-identity The parametrized curve
$$
\left( \sec\theta+\csc\theta,\  2\sqrt{2}\csc(2\theta) \right), \qquad \frac{10}{100} \le\theta\le\frac{142}{100}
$$
looks to the naked eye like a straight line.  The $y$-intercept is not $0$ and the slope is a number that I haven't tried to make sense out of (yet).  (The factor $2\sqrt{2}$ was chosen only to make the minimum values of the two coordinates equal to each other.)  The best-fitting straight line gives all residuals less than $0.08$, quite small!
Why?
 [/INST] Call $K=\csc\left(\theta\right)+\sec\left(\theta\right)$ and $W=a/\sin\left(2\theta\right)$ for some $a\gt0$ and $\theta\in\left]0,\pi/2\right[$ instead. You are interested in why $K$ and $W$ almost have a linear relationship. First, we know that $$K^2=\frac{1}{\sin^2\left(\theta\right)}+\frac{1}{\cos^2\left(\theta\right)}+\frac{2}{\sin\left(\theta\right)\cos\left(\theta\right)}\\=\frac{4a^2}{a^2\cdot 4\sin^2\left(\theta\right)\cos^2\left(\theta\right)}+\frac{4a}{a\cdot 2\sin\left(\theta\right)\cos\left(\theta\right)}=\frac{4W^2}{a^2}+\frac{4W}{a}\\\Longrightarrow K=\frac{2W}{a}\left(1+\frac{a}{W}\right)^{1/2}\\ \Longrightarrow \csc\left(\theta\right)+\sec\left(\theta\right)=\frac{2\sqrt{1+\sin(2\theta)}}{\sin\left(2\theta\right)}.$$ This is an exact identity. In the given interval, $\displaystyle \left|\sin\left(2\theta\right)\right| =\left|\frac{a}{W}\right|\lt 1$. We can invoke the binomial theorem: $$K=\frac{2}{\sin\left(2\theta\right)}\left[1+\frac{\sin\left(2\theta\right)}{2}-\frac{\sin^2\left(2\theta\right)}{8}+\text{higher order terms}\right]\approx \frac{2}{\sin\left(2\theta\right)}+1$$ as the other terms are going to zero. As said, this is almost a linear relationship. How good is it? Not much.
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