If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$,where –1 $$ \le $$ x $$ \le $$ 1, – 2 $$ \le $$ y $$ \le $$ 2, x $$ \le $$ $${y \over 2}$$ , then for all x, y, 4x² – 4xy cos $$\alpha $$ + y² is equal to :

If $$\alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$, $$\beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right)$$ where $$0 < \alpha ,\beta < {\pi \over 2}$$ , then $$\alpha $$ - $$\beta $$ is equal to :

Considering only the principal values of inverse functions, the set
A = { x $$ \ge $$ 0: tan^$$-$$1(2x) + tan^$$-$$1(3x) = $${\pi \over 4}$$}

All x satisfying the inequality (cot^–1 x)²– 7(cot^–1 x) + 10 > 0, lie in the interval :