Considering the principal values of inverse trigonometric functions, the value of the expression tan (2 sin ^-1 ( 2 √13 )-2 cos ^-1 ( 3 √10 ) ) is…

Let $\alpha = \sin^{-1}\left(\frac{2}{\sqrt{13}}\right)$α=sin−1(13​2​).

Then $\sin\alpha = \frac{2}{\sqrt{13}}$sinα=13​2​, $\cos\alpha = \frac{3}{\sqrt{13}}$cosα=13​3​, $\tan\alpha = \frac{2}{3}$tanα=32​. So $\tan(2\alpha) = \frac{2 \cdot \frac{2}{3}}{1 - \frac{4}{9}} = \frac{12}{5}$tan(2α)=1−94​2⋅32​​=512​.

Let $\gamma = \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)$γ=cos−1(10​3​).

Then $\cos\gamma = \frac{3}{\sqrt{10}}$cosγ=10​3​, $\sin\gamma = \frac{1}{\sqrt{10}}$sinγ=10​1​, $\tan\gamma = \frac{1}{3}$tanγ=31​. So $\tan(2\gamma) = \frac{2 \cdot \frac{1}{3}}{1 - \frac{1}{9}} = \frac{3}{4}$tan(2γ)=1−91​2⋅31​​=43​.

Therefore $\tan(2\alpha - 2\gamma) = \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}} = \frac{\frac{33}{20}}{\frac{56}{20}} = \frac{33}{56}$tan(2α−2γ)=1+512​⋅43​512​−43​​=2056​2033​​=5633​