Ta có:

a) 

$Có 0^{o} < 15^{o} < 90^{o} nên cos15^{o} > 0$

Lại có $sin^{2}15^{o} + cos^{2}15^{o} = 1$

=> $cos^{2}15^{o} = 1 - sin^{2}15^{o}= 1 - (\frac{\sqrt{6}-\sqrt{2}}{4})^{2} = \frac{2+\sqrt{3}}{4}$

=> $cos15^{o} = \frac{\sqrt{6}-\sqrt{2}}{4}$

$tan15^{o} = \frac{sin15^{o}}{cos15^{o}} = \frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}} = \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} = \frac{(\sqrt{6}-\sqrt{2}^{})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} = \frac{8-4\sqrt{3}}{4} = 2-\sqrt{3}$

Có $sin75^{o} = sin(90^{o}-15^{o}) = cos15^{o} = \frac{\sqrt{6}+\sqrt{2}}{4}$

$cos105^{o} = cos(180^{o} - 75^{o}) = -cos75^{o} = -cos(90^{o}-15^{o}) = -sin15^{o}$

Vậy $cos105^{o} = -\frac{\sqrt{6}-\sqrt{2}}{4}$

$tan165^{o} = tan(180^{o} - 15^{o} = -tan15^{o}$

Vậy $tan165^{o} = -2+\sqrt{3} $

b) 

Có $sin165^{o} = sin(180^{o} - 15^{o}) = sin15^{o} = \frac{\sqrt{6}-\sqrt{2}}{4}$

$cos165^{o} = cos(180^{o} - 15^{o}) = -cos15^{o} = -\frac{\sqrt{6}+\sqrt{2}}{4}$

Áp dụng vào $A = sin75^{o} . cos165^{o} + cos165^{o} . sin165^{o}$

$A = \frac{\sqrt{6}+\sqrt{2}}{4}.(-\frac{\sqrt{6}+\sqrt{2}}{4})+\frac{\sqrt{2}-\sqrt{6}}{4}.\frac{\sqrt{6}-\sqrt{2}}{4}$

A = -1