Answer
$\dfrac{3-2\sqrt{14}}{\sqrt 7+6\sqrt 2}$
Work Step by Step
Let us consider $p=\sin^{-1} \dfrac{\sqrt 3}{4}$ and $q=\cos^{-1} \dfrac{1}{3}$
This gives $\sin p=\dfrac{\sqrt 3}{4} \implies \cos p=\dfrac{\sqrt7}{4}$
and $\cos q=\dfrac{1}{3} \implies \sin q=\dfrac{2\sqrt 2}{3}$
Now, $\tan(\sin^{-1} \dfrac{3}{4}-\cos^{-1} \dfrac{1}{3} )=\tan (p- q)$
This implies that
$\tan (p- q)=\dfrac{\tan p-\tan q}{1+\tan p
tan q}$
This gives: $\dfrac{\tan p-\tan q}{1+\tan p
tan q}=\dfrac{\dfrac{3}{\sqrt 7}-2\sqrt 2}{1+\dfrac{6\sqrt 2}{\sqrt 7}}$
Hence, $\tan(\sin^{-1} \dfrac{3}{4}-\cos^{-1} \dfrac{1}{3} )=\dfrac{3-2\sqrt{14}}{\sqrt 7+6\sqrt 2}$
