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আমরা জানি, \(sec^2⁡θ-tan^2⁡θ=1 \)
বা, \((secθ+tanθ)(secθ-tanθ)=1 \)
বা, \((secθ+tanθ)×\cfrac{1}{√3}=1 \)
বা, \((secθ+tanθ)=√3\)
\(∴(secθ+tanθ)+(secθ-tanθ)\) \(=√3+\cfrac{1}{√3}\)
বা, \(secθ+tanθ+secθ-tanθ=\cfrac{3+1}{√3}\)
বা, \(2secθ=\cfrac{4}{√3}\)
বা, \(secθ=\cfrac{2}{√3} \)

এখন, \((secθ-tanθ)=\cfrac{1}{√3}\) সমীকরণে \(secθ =\cfrac{2}{√3}\) বসিয়ে পাই
\(\cfrac{2}{√3}-tanθ=\cfrac{1}{√3 }\)
বা, \(-tanθ=\cfrac{1}{√3}-\cfrac{2}{√3}\)
বা, \(-tanθ=-\cfrac{1}{√3}\)
বা, \(tanθ=\cfrac{1}{√3}\)

∴নির্ণেয় সমাধান, \(secθ=\cfrac{2}{√3}\) এবং \(tanθ=\cfrac{1}{√3}\)