In an isosceles right triangle $$PQR, \angle PRQ = 90^\circ$$. The points S and T are two trisection points of QR. The value of $$\tan(\angle SPT)$$ is

Let the isosceles sides be of the length 3x units each. Then, we can draw the triangle with given point as:

Angle SPT= B-A.

tanA= $$\ \frac{\ RT}{PR}=\ \frac{\ x}{3x}=\ \frac{\ 1}{3}\ $$

tanB= $$\ \frac{\ RS}{PR}=\ \frac{\ 2x}{3x}=\ \frac{\ 2}{3}\ $$

So, $$\tan\left(B-A\right)=\ \ \frac{\ \tan B-\tan\ A}{1+\tan\ A\tan\ B}\ $$.

=> $$\ \tan\left(B-A\right)=\ \ \frac{\ \ \frac{\ 2}{3}-\ \frac{\ 1}{3}}{1+\ \frac{\ 2}{9}}$$

=>$$\ \tan\left(B-A\right)=\ \ \frac{\ \ \frac{\ 1}{3}}{\ \ \frac{\ 11}{9}}=\ \frac{\ 3}{11}$$