Step-by-step explanation:
[tex]\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}[/tex]
[tex]\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}[/tex]
[tex]\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}[/tex]
[tex]\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}[/tex]
[tex]\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}[/tex]
[tex]\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}[/tex]
Taking sin²θ common in both numerator & denominator, We get :
[tex]\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}[/tex]
[tex]\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}[/tex]
[tex]\mathsf{\implies \dfrac{1 -cot^2\theta}{1 + cot^2\theta}}[/tex]
[tex]\mathsf{Given :\;cot\theta = \dfrac{1}{\sqrt{5}}}[/tex]
[tex]\mathsf{\implies \dfrac{1 - \left(\dfrac{1}{\sqrt{5}}\right)^2}{1 + \left(\dfrac{1}{\sqrt{5}}\right)^2}}[/tex]
[tex]\mathsf{\implies \dfrac{1 - \dfrac{1}{5}}{1 + \dfrac{1}{5}}}[/tex]
[tex]\mathsf{\implies \dfrac{\dfrac{5 - 1}{5}}{\dfrac{5 + 1}{5}}}[/tex]
[tex]\mathsf{\implies \dfrac{5 - 1}{5 + 1}}[/tex]
[tex]\mathsf{\implies \dfrac{4}{6}}[/tex]
[tex]\mathsf{\implies \dfrac{2}{3}}[/tex]
Hence, option (a) 2/3 is your correct answer.
