91Ó°ÊÓ

Let's call the angle whose cosine is $$\frac{2}{3}$$ as $$A$$, so we have $$\cos A = \frac{2}{3}$$. Now, let's create a right triangle with angle $$A$$ such that its adjacent side is 2 and its hypotenuse is 3. Applying the Pythagorean theorem, we find the length of the opposite side to angle $$A$$: \[ b = \sqrt{3^2 - 2^2} = \sqrt{5} \] Now let's do the same for $$\tan ^{-1} 3$$. Let's call the angle whose tangent is $$3$$ as $$B$$, so we have $$\tan B = 3$$. We can create a right triangle with angle $$B$$ such that its opposite side is 3 and its adjacent side is 1. The hypotenuse of this triangle can be found by the Pythagorean theorem: \[ c = \sqrt{1^2 + 3^2} = \sqrt{10} \]

Now, we are supposed to find the cosine of the sum of angles $$A$$ and $$B$$, i.e., $$\cos(A+B)$$. Using the sum of angles formula for cosine: \[ \cos(A+B) = \cos A \cos B - \sin A \sin B \] In order to find the values of $$\cos A, \cos B, \sin A,$$ and $$\sin B$$, we use the lengths of the sides of the right triangles we created in the previous step For triangle with angle $$A$$: \[ \cos A = \frac{2}{3}, \sin A = \frac{\sqrt{5}}{3} \] For the triangle with angle $$B$$: \[ \cos B = \frac{1}{\sqrt{10}}, \sin B = \frac{3}{\sqrt{10}} \] Plugging these values into the formula for $$\cos(A+B)$$: \[ \cos(A+B) = \left(\frac{2}{3}\right) \left(\frac{1}{\sqrt{10}}\right) - \left(\frac{\sqrt{5}}{3}\right) \left(\frac{3}{\sqrt{10}}\right) \]

Now, let's simplify the expression for $$\cos(A+B)$$: \[ \cos(A+B) = \frac{2}{3\sqrt{10}} - \frac{3\sqrt{5}}{3\sqrt{10}} \] Combining the terms gives the final answer: \[ \cos(A+B) = \frac{2 - 3\sqrt{5}}{3\sqrt{10}} \] So, the value of the given expression $$\cos \left(\cos ^{-1} \frac{2}{3}+\tan ^{-1} 3\right)$$ is $$\frac{2 - 3\sqrt{5}}{3\sqrt{10}}$$.