How do you prove #sec(tan^-1(sqrt3/3))#?

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Answer 1 · 2018-04-15T09:35:26

#sec(tan^(-1)(sqrt3/3))=(2sqrt3)/3#

Let #tan^(-1)(sqrt3/3)=A# then #tanA=sqrt3/3#.

#rarrsecA=sqrt(1+tan^2A)=sqrt(1+(sqrt3/3)^2)=(2sqrt3)/3#

#rarrA=sec^(-1)((2sqrt3)/3)#

Now, #sec(tan^(-1)(sqrt3/3))=sec(sec^(-1)((2sqrt3)/3))=(2sqrt3)/3#

Answer 2 · 2018-04-18T05:48:46

#color(indigo)(=> 2 / sqrt 3 " or " (2 sqrt 3) / 3#

#sec (tan^-1 (sqrt3 / 3))#

#=> sec (tan^-1 (1/sqrt3))#

www.math-only-math.com/images/trigonometrical-ratios-table.png

#=> sec (tan^-1( tan (pi/6)), color(violet)(" as "tan (pi/6) = 1 / sqrt3 " from table above"#

#=> sec (pi/6) = 1/ cos (pi/6) = 1 / (sqrt 3/2)#

#color(indigo)(=> 2 / sqrt 3 " or " (2 sqrt 3) / 3#