Question

How do you evaluate `tan(arccos(-9/10))`?

Answer 1

`-sqrt19/9.`

Explanation:

We have, `color(red)((1)cos^-1(-x)=pi-cos^-1x`
`color(red)((2)tan(pi-theta)=-tantheta`
`color(red)((3)cos^-1x=tan^-1((sqrt(1-x^2)/x)`
`color(red)((4)tan(tan^-1x)=x`

`tan(cos^-1(-9/10))=tan(pi-cos^-1(9/10)).to`Apply(1)
`=-tan(cos^-1(9/10))...........to`Apply(2)
`=-tan(tan^-1(sqrt(1-(9/10)^2)/(9/10)))...to`Apply(3)
`=-tan(tan^-1(sqrt(1-(81/100))/(9/10)))`
`=-tan(tan^-1((sqrt(100-81)/10)/(9/10)))`
`=-tan(tan^-1(sqrt19/9))`

`=-sqrt19/9....to` Apply(4)