Let $$\cos \left( {\alpha + \beta } \right) = {4 \over 5}$$ and $$\sin \,\,\,\left( {\alpha - \beta } \right) = {5 \over {13}},$$ where $$0 \le \alpha ,\,\beta \le {\pi \over 4}.$$
Then $$tan\,2\alpha $$ =

Let A and B denote the statements

A: $$\cos \alpha + \cos \beta + \cos \gamma = 0$$

B: $$\sin \alpha + \sin \beta + \sin \gamma = 0$$

If $$\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - {3 \over 2},$$ then:

If $$0 < x < \pi $$ and $$\cos x + \sin x = {1 \over 2},$$ then $$\tan x$$ is :

If $$u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $$

then the difference between the maximum and minimum values of $${u^2}$$ is given by :