If  $3Ta{n}^4\alpha -10Ta{n}^2\alpha +3=0$ then principal values of ‘$\alpha$’ are

Solution:

$3{\tan }^4\alpha -10{\tan }^2\alpha +3=0$

$\Rightarrow 3{\tan }^4\alpha -9{\tan }^2\alpha -{\tan }^2\alpha +3=0$

$\Rightarrow (3{\tan }^2\alpha -1)({\tan }^2\alpha -3)=0$

$\Rightarrow {\tan }^2\alpha =\frac{1}{3}={\left(\frac{{\displaystyle 1}}{{\displaystyle \sqrt{3}}}\right)}^2$$;{\tan }^2\alpha =3={\left(\sqrt{3}\right)}^2$

$\therefore \alpha =\pm {30}^0,\pm {60}^0$