Question

Evaluate the following:

`tan  1/2(cos^-1  sqrt5/3)`

Answer 1

Let, `cos^-1  sqrt5/3=theta`

`=> costheta=sqrt5/3`

`=>2cos^2  theta/2-1=sqrt5/3`

`=>cos^2  theta/2=(3+sqrt5)/6`

`=>theta/2=cos^-1(sqrt((3+sqrt5)/6))`

`=tan^-1((sqrt(1-(sqrt((3+sqrt5)/6))^2))/(sqrt((3+sqrt5)/6)))`

`=tan^-1(sqrt(1-(3+sqrt5)/6)/sqrt(3+sqrt5/6))`

`=tan^-1((sqrt((3-sqrt5)/6))/(sqrt((3+sqrt5)/6)))`

`=tan^-1(sqrt((3-sqrt5)/(3+sqrt5)))`

`=tan^-1(sqrt(((3-sqrt5)(3-sqrt5))/((3+sqrt5)(3-sqrt5))))`

`=tan^-1(sqrt((3-sqrt5)^2/(9-5)))`

`=tan^-1((3-sqrt5)/2)`

i. e. , `1/2(cos^-1  sqrt5/3)=tan^-1  ((3-sqrt5)/2)`

`=>tan  1/2(cos^-1  sqrt5/3)=tan[tan^-1((3-sqrt5)/2)]`

`thereforetan  1/2(cos^-1  sqrt5/3)=(3-sqrt5)/2`