দেওয়া আছে, $$ \begin{aligned} \tan \left(\sin ^{-1} \sqrt{1-x^{2}}\right)&=\sin \left(\cos ^{-1} \frac{1}{\sqrt{5}}\right) \end{aligned} $$ ধরি, $\sin^{-1} \sqrt{1-x^2} = \theta$, তাহলে $\sin \theta = \sqrt{1-x^2}$। এখান থেকে পাই, $\tan \theta = \frac{\sqrt{1-x^2}}{x}$। আবার, $\cos^{-1} \frac{1}{\sqrt{5}} = \phi$ ধরলে, $\cos \phi = \frac{1}{\sqrt{5}}$। এখান থেকে পাই, $\sin \phi = \sqrt{1 - (\frac{1}{\sqrt{5}})^2} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$। সুতরাং, $$ \begin{aligned} \frac{\sqrt{1-x^2}}{x}&=\frac{2}{\sqrt{5}} \end{aligned} $$ উভয়পক্ষকে বর্গ করে পাই, $$ \begin{aligned} \frac{1-x^2}{x^2}&=\frac{4}{5} \end{aligned} $$ $$ \begin{aligned} 5-5x^2&=4x^2 \end{aligned} $$ $$ \begin{aligned} 9x^2&=5 \end{aligned} $$ $$ \begin{aligned} x^2&=\frac{5}{9} \end{aligned} $$ $$ \begin{aligned} x&=\pm \frac{\sqrt{5}}{3} \end{aligned} $$