If $$A=$$$$tan$$−$$1$$⁡$$x32k$$−x and $$B=$$$$tan$$−$$1$$⁡$$2x$$−$$kk3$$, then the value of $$A-B$$ in degrees, is

We find that $$tan$$⁡$$(A$$−$$B)=$$$$tan$$⁡A−$$tan$$⁡$$B1+$$$$tan$$⁡Atan⁡$$B=x32k$$−x−$$2x$$−$$kk31+x32k$$−x⋅$$2x$$−$$kk3=13$$⇒ A−$$B=30$$∘

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If $A = \tan^{- 1} (\frac{x \sqrt{3}}{2 k - x})$ and $B = \tan^{- 1} (\frac{2 x - k}{k \sqrt{3}})$ , then the value of $A - B$ in degrees, is

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Solution

We find that
$\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\frac{x \sqrt{3}}{2 k - x} - \frac{2 x - k}{k \sqrt{3}}}{1 + \frac{x \sqrt{3}}{2 k - x} \cdot \frac{2 x - k}{k \sqrt{3}}} = \frac{1}{\sqrt{3}}$
$\Rightarrow A - B = 30^{\circ}$

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