Question

If `sin^( 2)( \theta )+ 3cos^( 2)( \theta )= 4` then show that `tan(\theta )= \pm ( 1)/( sqrt ( 3) )` ?

Answer 1

Yes, it works.
See explanation.

Explanation:

We have:
`sin^2(theta) + 3cos^2(theta)=4`

which really is just:
`sin^2(theta)+cos^2(theta) + 2cos^2(theta)=4`

and using the identity `sin^2(x) + cos^2(x)=1` for all `x`, we get:
`1+2cos^2(theta)=4` (eq.A)

We carry on and simplify eq.A until we get an expression for `cos(theta)`:
`2cos^2(theta)=3`
`cos^2(theta)=3/2`
`cos(theta)=+-sqrt(3/2)`

We also notice that `cos^2(theta) =1-sin^2(theta)`, so (eq.A) becomes:
`1+2cos^2(theta)=1+2(1-sin^2(theta))=4`
i.e.
`3-2sin^2(theta)=4`
i.e.
`sin^2(theta)=-1/2`
`sin(theta)= +- sqrt(1/2)` (it should really be 'minus-plus' instead but the symbol does not exist here in Socratic).

So, now it is simply a matter of plugging these in the `tan` function:
`tan(theta)=sin(theta)/cos(theta)`
`=(+- sqrt(1/2))/(+- sqrt(3/2))`
`=+-(1/sqrt(3))`
`=+-(sqrt 3)/3` (it is more conventional to write it like this without the square-root in the denominator).