Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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

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


`=> 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`