How to solve theta = tan inverse ____any value??

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In the experiment to verify inverse square law with deflection Magnetometer, the value of $\frac{tan θ_{A}}{tan θ_{B}}$ will come out as :

A = [\begin{matrix} sin θ & tan θ tan θ & sin θ \end{matrix}]

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

2 {t a n}^{- 1} (\sqrt{\frac{a - b}{a + b}} t a n \frac{θ}{2}) = {c o s}^{- 1} \frac{a c o s θ + b}{a + b c o s θ}

a = t a n θ + c o t θ

b = t a n θ - c o t θ

a^{2} - b^{2} + (\frac{a + b}{t a n θ})

[\begin{matrix} sec θ & - tan θ - tan θ & sec θ \end{matrix}]

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