[tex]\text{Given that,}\\\\\cot A = -\dfrac{9}{\sqrt{40}}\\\\\\\implies \tan A = -\dfrac{\sqrt{40}}9\\\\\\\implies \tan^2 A = \dfrac{40}{81}\\\\\\\implies \sec^2 A -1 = \dfrac{40}{81}\\\\\implies \sec^2 A = 1 + \dfrac{40}{81}\\\\\implies \sec^2 A = \dfrac{121}{81}\\\\\implies \sec A = \pm \sqrt{\dfrac{121}{81}}= \pm\dfrac{11}{9}\\\\\\\text{Since the angle A is in quadrant II, the value of sec A will be negative.}\\\\\\\text{Hence,}~ \sec A=-\dfrac{11}{9}[/tex]
