Question

How do you evaluate `tan(csc^-1(sqrt7/2))` without a calculator?

Answer 1

`tan(csc^(-1)(sqrt(7)/2)) = (2sqrt(3))/3`

Explanation:

Consider a triangle with sides `2`, `sqrt(3)` and `sqrt(7)`.

It is a right angled triangle, since:

`2^2+(sqrt(3))^2 = 4+3 = 7 = (sqrt(7))^2`

Since `csc theta = 1/sin theta = "hypotenuse"/"opposite"`, and `tan theta = "opposite"/"adjacent"` we have:

`"hypotenuse" = sqrt(7)`

`"opposite" = 2`

`"adjacent" = sqrt(3)`

`tan(csc^(-1)(sqrt(7)/2)) = 2/sqrt(3) = (2sqrt(3))/3`

Answer 2

`=(2sqrt3)/3`

Explanation:

Let `theta=csc^-1(sqrt(7)/2)`
or
`csctheta=sqrt7/2`
or
`1/sintheta=sqrt7/2`
or
`sintheta=2/sqrt7`
or
`theta=sin^-1(2/sqrt7)`
Hence
`p/h=2/sqrt7`
`costheta=b/h`
or
`costheta=sqrt(h^2-p^2)/h`
or
`costheta=(sqrt((sqrt7)^2-2^2))/(sqrt7)`
or
`costheta=sqrt(7-4)/sqrt7`
or
`costheta=sqrt3/sqrt7`
Therefore
`tantheta=sintheta/costheta`
or
`tan theta=(2/sqrt7)/((sqrt3)/sqrt7)`
or
`tan theta=2/sqrt3`
or
`tan(csc^-1(sqrt(7)/2))=2/sqrt3`
or
`tan(csc^-1(sqrt(7)/2))=(2sqrt3)/3`

Answer 3

`tan(csc^-1(sqrt7/2))`

`=tan(cot^-1(sqrt((sqrt7/2)^2-1)))`

`=tan(cot^-1sqrt(7/4-1))`

`=tan(cot^-1sqrt(3/4))`

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