If $\text{cos A} = \dfrac{1}{\sqrt2}$ and $\text{sin B} = \dfrac{\sqrt3}{2}$, the value of $\dfrac{\text{tan B} - \text{tan A}}{1 + \text{tan A tan B}}$ is :

1. $2 - {\sqrt3}$
2. $2 + {\sqrt3}$
3. ${\sqrt3} - 2$
4. ${\sqrt3} + 2$

Given:

cos $A = \dfrac{1}{\sqrt2}$

i.e., $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{1}{\sqrt2}$

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

In Δ AMO,

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

⇒ ($\sqrt{2}$x)² = (x)² + MO²

⇒ 2x² = x² + MO²

⇒ MO² = 2x² - x²

⇒ MO² = x²

⇒ MO = $\sqrt{\text{x}^2}$

⇒ MO = x

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

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

And,

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

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

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

In Δ BXY,

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

⇒ (2y)² = ($\sqrt{3}$y)² + BX²

⇒ 4y² = 3y² + BX²

⇒ BX² = 4y² - 3y²

⇒ BX² = y²

⇒ BX = $\sqrt{\text{y}^2}$

⇒ BX = y

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

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

Now,

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

Hence, option 1 is the correct option.