দেওয়া আছে, $3 \sec^4 \theta - 10 \sec^2 \theta + 8 = 0$। ধরি, $\sec^2 \theta = x$। $$ \begin{aligned} 3x^2 - 10x + 8 &= 0 \\ \Rightarrow 3x^2 - 6x - 4x + 8 &= 0 \\ \Rightarrow 3x(x-2) - 4(x-2) &= 0 \\ \Rightarrow (3x-4)(x-2) &= 0 \end{aligned} $$ সুতরাং, $x = 2$ অথবা $x = \frac{4}{3}$। ১) $\sec^2 \theta = 2 \Rightarrow 1 + \tan^2 \theta = 2 \Rightarrow \tan^2 \theta = 1 \Rightarrow \tan \theta = \pm 1$। ২) $\sec^2 \theta = \frac{4}{3} \Rightarrow 1 + \tan^2 \theta = \frac{4}{3} \Rightarrow \tan^2 \theta = \frac{1}{3} \Rightarrow \tan \theta = \pm \frac{1}{\sqrt{3}}$।