How do you simplify $sec (arctan 2)$ $sec(arctan 2)$ ?

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Answer 1 · 2018-03-17T05:47:45

$\sqrt{5}$ $sqrt5$ .

Let, $a r c tan 2 = θ, θ \in (- \frac{π}{2}, \frac{π}{2})$ $arc tan2=theta, theta in (-pi/2,pi/2)$ .

$:. tantheta=2>0 rArr theta in (0,pi/2)$ .

Hence, $sec(arc tan2)$ ,

$=sectheta$ ,

$=+sqrt(sec^2theta)......[because, theta in (0,pi/2)]$ ,

$=+sqrt(1+tan^2theta)$ ,

$=+sqrt(1+2^2)$ .

$rArr sec(arc tan2)=sqrt5$ .

How do you simplify sec(arctan 2)sec(arctan2)sec(arctan 2)?