Question

If `cos(50)=a`, then how do you express `tan(130)` in terms of a?

Answer 1

One method

Assume `a=cos50^0`
`tan(130^0) = sin(130^0)/cos(130^0)`
`sin(130^0)=sin(90^0+40^0)`
Expand using `sin(A+B)` formula
This will give you
`sin(130^0) = cos(40^0)`
`cos(40^0) = sin(50^0)` therefore
`sin(130^0) = sin(50^0)`
`sin(130^0) = +sqrt(1-cos^2(50^0))`
`sin(130^0) =sqrt(1-a^2)`
Similarly for `cos(130^0)` we get
`cos(130^0) = -sin(40^0)`
`cos (50^0)=sin(40^0)` hence
`cos(130^0) = -cos(50^0)`
Combining them gives

`tan(130^0) = (-sqrt(1-a^2))/a`

Answer 2

You can use some identities to make this easier.

`cos(50^o) = cos(-50^o) = -cos(130^o)`

due to

`cos(x) = cos(-x)`
`cos(x) = -cos(x+pi)`

Thus, you have:

`tan(130^o) = (sin(130^o))/(cos(130^o)) = (-sin(130^o))/(-cos(130^o))`

Now, as for determining `-sin(130^o)`... Note that `sin^2(130^o) + cos^2(130^o) = 1`. Therefore:

`sin(130^o) = sqrt(1-cos^2(130^o))`

`=> -sin(130^o) = -sqrt(1-cos^2(130^o))`

Finally, you get:

`color(blue)(tan(130^o)) = (-sin(130^o))/(-cos(130^o)) = (-sqrt(1-(cos(130^o))^2))/(-cos(130^o))`

`= (-sqrt(1-(-cos(50^o))^2))/(cos(50^o))`

`= (-sqrt(1-cos^2(50^o)))/(cos(50^o))`

`= color(blue)((-sqrt(1-a^2))/(a))`