Question

Given sin0=6/11 and sec 0<0 find cos0 and tan0 (picture provided)

Answer 1

`cos(0\°)=-(√(85))/(11)`

`tan(0\°) = -(6)/(√(85))\\\\`

Explanation:

We know by definition that:

`cos ^ 2(x) = 1-sin ^2(x)`

We also know that:

`tan(x) = (sin(x))/(cos(x))`

`sec(x) = (1)/(cos(x))`

Then we can use these identities to solve the problem

if
`sin(0\°) = (6)/(11)` then
`cos^2(0\°) = 1-((6)/(11))^2`

`cos(0\°) = \±\sqrt{(85)/(121)}\\\\cos(0\°)=\±(√(85))/(11)`

As
`sec(x) <0` then
`cos(x) <0`. Therefore we take the negative root:

`cos(0\°)=-(√(85))/(11)`

Now that we know
`sin(0\°)` and
`cos(0\°)` we can find
`tan(0\°)`

`tan(0\°) = (sin(0\°))/(cos(0\°))\\\\tan(0\°) = ((6)/(11))/((-√(85))/(11) )\\\\tan(0\°) = -(6)/(√(85))\\\\`