Question #8908d

Question

Question #8908d

1 Answer

Impact of this question

1226 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-24T13:15:59

$sec A = \pm \frac{\sqrt{a^{2} + b^{2}}}{b}$ $secA=+-sqrt(a^2+b^2)/b$ .

Explanation:

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

So, we do it this way:-

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

$= \frac{a^{2} + b^{2}}{a^{2}}$ $=(a^2+b^2)/a^2$

$:. secA=+-sqrt(a^2+b^2)/b$ .

Science

Math

Humanities

... and beyond

Question #8908d

1 Answer

Explanation:

Related questions

Impact of this question