দেওয়া আছে, $3 \sec^4 \theta - 10 \sec^2 \theta + 8 = 0$। আমরা জানি, $\sec^2 \theta = 1 + \tan^2 \theta$। $$ \begin{aligned} 3(1+\tan^2 \theta)^2 - 10(1+\tan^2 \theta) + 8 &= 0 \\ \Rightarrow 3(1 + 2\tan^2 \theta + \tan^4 \theta) - 10 - 10\tan^2 \theta + 8 &= 0 \\ \Rightarrow 3 + 6\tan^2 \theta + 3\tan^4 \theta - 2 - 10\tan^2 \theta &= 0 \\ \Rightarrow 3\tan^4 \theta - 4\tan^2 \theta + 1 &= 0 \\ \Rightarrow 3\tan^4 \theta - 3\tan^2 \theta - \tan^2 \theta + 1 &= 0 \\ \Rightarrow 3\tan^2 \theta(\tan^2 \theta - 1) - 1(\tan^2 \theta - 1) &= 0 \\ \Rightarrow (3\tan^2 \theta - 1)(\tan^2 \theta - 1) &= 0 \end{aligned} $$ সুতরাং, $\tan^2 \theta = 1$ বা $\tan^2 \theta = \frac{1}{3}$। অর্থাৎ, $\tan \theta = \pm 1$ বা $\tan \theta = \pm \frac{1}{\sqrt{3}}$।