Question

If $3{\sec }^4\theta +8=10{\sec }^2\theta$, find the value of $\tan \theta$.

Answer 1

Given $3{\text{sec}}^4\theta +8=10{\text{sec}}^2\theta$

$⟹3({\text{sec}}^2\theta {)}^2-10{\text{sec}}^2\theta +8=0$

$⟹{\text{sec}}^2\theta =\frac{10\pm \sqrt{100-96}}{6}=\frac{10\pm 2}{6}=2,\frac{4}{3}$

$⟹{\tan }^2\theta =1,\frac{1}{3}⟹\tan \theta =\pm 1,\pm \frac{1}{\sqrt{3}}$