# Question #8908d

Jul 24, 2016

$\sec A = \pm \frac{\sqrt{{a}^{2} + {b}^{2}}}{b}$.

#### Explanation:

Though the Question is not clearly placed, but as I understand, it asks to find the value of $\sec A$, given the value of $\cot A = \frac{a}{b}$.

So, we do it this way:-

$\cot A = \frac{a}{b} \Rightarrow \tan A = \frac{b}{a} \Rightarrow {\sec}^{2} A = 1 + {\tan}^{2} A = 1 + {b}^{2} / {a}^{2}$

$= \frac{{a}^{2} + {b}^{2}}{a} ^ 2$

$\therefore \sec A = \pm \frac{\sqrt{{a}^{2} + {b}^{2}}}{b}$.