Suppose we know $x \sec \theta = a$ and $x \operatorname{cosec} \theta = b$.
Is it possible to derive an expression for $x$ and $\theta$? I know that we can find $\theta$ (and then $x$) from an intersection of the plots, but is a trigonometric expression possible?
I have spent many days on this problem, but am unable to find a solution. Could the experts here please guide me if/when free? My knowledge of trigonometric identities is decent, so I would gratefully embrace even a hint.
[Having stopped formal education a decade ago, I beg you to not doubt this as a request for homework help.]
\begin{align} \frac{x}{\cos\left( \theta\right) } & =a\tag{1}\\ \frac{x}{\sin\left( \theta\right) } & =b\tag{2} \end{align} Hence \begin{align*} \cos\left( \theta\right) & =\frac{x}{a}\\ \sin\left( \theta\right) & =\frac{x}{b} \end{align*} Since $$ 1=\cos^{2}\left( \theta\right) +\sin^{2}\left( \theta\right) $$ Then \begin{align*} 1 & =\left( \frac{x}{a}\right) ^{2}+\left( \frac{x}{b}\right) ^{2}\\ 1 & =x^{2}\left( \frac{1}{a^{2}}+\frac{1}{b^{2}}\right) \\ & =x^{2}\left( \frac{b^{2}+a^{2}}{a^{2}b^{2}}\right) \end{align*} Solving for $x$ gives \begin{align} x^{2} & =\frac{a^{2}b^{2}}{b^{2}+a^{2}}\nonumber\\ x & =\pm\frac{ab}{\sqrt{a^{2}+b^{2}}}\tag{3} \end{align} To find $\theta$, we divide (1) by (2) \begin{align*} \frac{\frac{x}{\cos\left( \theta\right) }}{\frac{x}{\sin\left( \theta\right) }} & =\frac{a}{b}\\ \tan\theta & =\frac{a}{b}\\ \theta & =\arctan\left( \frac{a}{b}\right) \end{align*} This assumes $a>0,b>0$.