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

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Answer 1 · 2018-03-17T05:47:45

# sqrt5#.

Let, #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#.