Question

find q if x = ( sec q + tan q) and y = b( sec q - tan q)

Answer 1

find q if $x = (\sec q + \tan q)$ and $y = b(\sec q - \tan q)$

Solution

Multiply $x$ by $y$ :

$$\begin{array}{l} x y = b (\sec q + \tan q) (\sec q - \tan q) = b (\sec q ^ {2} - \tan q ^ {2}) = b \left(\frac {1}{\cos q ^ {2}} - \tan q ^ {2}\right) \\ = \left| \frac {1}{\cos q ^ {2}} = 1 + \tan q ^ {2} \right| = b * 1 = b \rightarrow b = x y \\ x = (\sec q + \tan q) = \frac {1}{\cos q} + \frac {\sin q}{\cos q} = \frac {1 + \sin q}{\cos q} \\ \end{array}$$

Let's use $\left\{ \begin{array}{l}\cos \left(\frac{\pi}{2} -q\right) = \sin q\\ \sin \left(\frac{\pi}{2} -q\right) = \cos q \end{array} \right.$ . Then

$$x = \frac {1 + \sin q}{\cos q} = \frac {1 + \cos \left(\frac {\pi}{2} - q\right)}{\sin \left(\frac {\pi}{2} - q\right)} = \cot \left(\frac {\frac {\pi}{2} - q}{2}, \text { because of } \cot \frac {q}{2} = \frac {1 + \cos q}{\sin q} \right.$$

So

$$x = \cot \left(\frac {\pi}{4} - \frac {q}{2}\right)\rightarrow \left(\frac {\pi}{4} - \frac {q}{2}\right) = \cot^ {- 1} x \rightarrow q = 2 \left(\frac {\pi}{4} - \cot^ {- 1} x\right) = \frac {\pi}{2} - 2 \cot^ {- 1} x.$$

Answer: $\frac{\pi}{2} - 2\cot^{-1}x$