If $\text{cos A} = \dfrac{1}{2}$ and $\text{sin B} = \dfrac{1}{\sqrt2}$, find the value of : $\dfrac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}}$. Here angles A and B are from different right triangles.

Given:

cos $A = \dfrac{1}{2}$

i.e., $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{1}{2}$

∴ If length of AM = 1x unit, length of AO = 2x unit.

In Δ AMO,

⇒ AO² = AM² + MO² (∵ AC is hypotenuse)

⇒ (2x)² = (1x)² + MO²

⇒ 4x² = 1x² + MO²

⇒ MO² = 4x² - 1x²

⇒ MO² = 3x²

⇒ MO = $\sqrt{3\text{x}^2}$

⇒ MO = $\sqrt{3}$ x

tan $A = \dfrac{Perpendicular}{Base}$

$= \dfrac{OM}{MA} = \dfrac{\sqrt{3} x}{1x} = \dfrac{\sqrt{3}}{1} = \sqrt{3}$

And,

sin $B = \dfrac{1}{\sqrt{2}}$

i.e., $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{1}{\sqrt{2}}$

∴ If length of XY = y unit, length of YB = y $\sqrt{2}$ unit.

In Δ BXY,

⇒ YB² = YX² + BX² (∵ AC is hypotenuse)

⇒ ($\sqrt{2}$y)² = (y)² + BX²

⇒ 2y² = y² + BX²

⇒ BX² = 2y² - y²

⇒ BX² = y²

⇒ BX = $\sqrt{\text{y}^2}$

⇒ BX = y

tan $B = \dfrac{Perpendicular}{Base}$

$= \dfrac{YX}{XB} = \dfrac{y}{y}= 1$

Now,

$\dfrac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}}\\[1em] = \dfrac{{\sqrt{3} - 1}}{1 + {\sqrt{3}}}\\[1em] = \dfrac{(\sqrt{3} - 1) \times (1 - \sqrt{3})}{(1 + \sqrt{3}) \times (1 - \sqrt{3})}\\[1em] = \dfrac{(\sqrt{3} - 3 - 1 + \sqrt{3})}{1^2 - \sqrt{3}^2}\\[1em] = \dfrac{2 \sqrt{3} - 4}{1 - 3}\\[1em] = \dfrac{2 (\sqrt{3} - 2)}{-2}\\[1em] = \dfrac{\cancel{2} (\sqrt{3} - 2)}{-\cancel{2}}\\[1em] = -\sqrt{3} + 2\\[1em] = 2 - \sqrt{3}$

Hence, $\dfrac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}} = 2 - \sqrt{3}$.