If $\text{sin A} = \dfrac{\sqrt3}{2}$ and $\text{cos B} = \dfrac{\sqrt3}{2}$, find the value of :

$\dfrac{\text{tan A} - \text{tan B}}{\text{1 + tan A tan B}}$.

Given:

$\text{sin A} = \dfrac{\sqrt3}{2}$

i.e., $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{\sqrt3}{2}$

∴ If length of OM = $\sqrt{3}$ x unit, length of AO = 2x unit.

In Δ AMO,

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

⇒ (2x)² = ($\sqrt{3}$x)² + AM²

⇒ 4x² = 3x² + AM²

⇒ AM² = 4x² - 3x²

⇒ AM² = x²

⇒ AM = $\sqrt{\text{x}^2}$

⇒ AM = x

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

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

And,

cos $B = \dfrac{\sqrt3}{2}$

i.e. $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{\sqrt3}{2}$

∴ If length of XB = $\sqrt{3}$ y unit, length of YB = 2y unit.

In Δ BXY,

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

⇒ (2y)² = ($\sqrt{3}$y)² + XY²

⇒ 4y² = 3y² + XY²

⇒ XY² = 4y² - 3y²

⇒ XY² = y²

⇒ XY = $\sqrt{\text{y}^2}$

⇒ XY = y

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

$= \dfrac{XY}{BX} = \dfrac{y}{\sqrt{3}y} = \dfrac{1}{\sqrt{3}}$

Now,

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

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