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Question
Evaluate the following:
`tan 1/2(cos^-1 sqrt5/3)`
Solution
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`
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