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

Feb 18, 2018

Yes, it works.
See explanation.

Explanation:

We have:
${\sin}^{2} \left(\theta\right) + 3 {\cos}^{2} \left(\theta\right) = 4$

which really is just:
${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) + 2 {\cos}^{2} \left(\theta\right) = 4$

and using the identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ for all $x$, we get:
$1 + 2 {\cos}^{2} \left(\theta\right) = 4$ (eq.A)

We carry on and simplify eq.A until we get an expression for $\cos \left(\theta\right)$:
$2 {\cos}^{2} \left(\theta\right) = 3$
${\cos}^{2} \left(\theta\right) = \frac{3}{2}$
$\cos \left(\theta\right) = \pm \sqrt{\frac{3}{2}}$

We also notice that ${\cos}^{2} \left(\theta\right) = 1 - {\sin}^{2} \left(\theta\right)$, so (eq.A) becomes:
$1 + 2 {\cos}^{2} \left(\theta\right) = 1 + 2 \left(1 - {\sin}^{2} \left(\theta\right)\right) = 4$
i.e.
$3 - 2 {\sin}^{2} \left(\theta\right) = 4$
i.e.
${\sin}^{2} \left(\theta\right) = - \frac{1}{2}$
$\sin \left(\theta\right) = \pm \sqrt{\frac{1}{2}}$ (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:
$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$
$= \frac{\pm \sqrt{\frac{1}{2}}}{\pm \sqrt{\frac{3}{2}}}$
$= \pm \left(\frac{1}{\sqrt{3}}\right)$
$= \pm \frac{\sqrt{3}}{3}$ (it is more conventional to write it like this without the square-root in the denominator).