Given:

2cos²θ – 5cosθ + 2 = 0

Concept used:

Using the concept of trigonometric ratios with respect to standard angles.

Calculation:

2cos²θ – 5cosθ + 2 = 0

⇒ 2cos²θ – 4cosθ – cosθ + 2 = 0

⇒ 2cosθ(cosθ – 2) – 1(cosθ – 2) = 0

⇒ (cosθ – 2)(2cosθ – 1) = 0

⇒ (cosθ – 2) = 0 or (2cosθ – 1) = 0

⇒ cosθ = 2 or cosθ = 1/2

cosθ ≠ 2, because maximum value of cosθ is 1.

So, cosθ = 1/2

⇒ cosθ = cos60°      [0° < θ < 90°]

⇒ θ = 60° 

\(\sqrt {\frac{{1{\rm{\;}} + {\rm{\;ta}}{{\rm{n}}^2}{\rm{θ }}}}{{1{\rm{\;}} - {\rm{\;co}}{{\rm{t}}^2}{\rm{θ }}}}} \)

\( \Rightarrow \sqrt {\frac{{1{\rm{\;}} + {\rm{\;ta}}{{\rm{n}}^2}60^\circ }}{{1{\rm{\;}} - {\rm{\;co}}{{\rm{t}}^2}60^\circ }}} \)

\( \Rightarrow \sqrt {\frac{{1{\rm{\;}} + {\rm{\;}}{{\left( {\sqrt 3 } \right)}^2}}}{{1{\rm{\;}} - {\rm{\;}}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}} \)

\( \Rightarrow \sqrt {\frac{{1{\rm{\;}} + {\rm{\;}}3}}{{1{\rm{\;}} - {\rm{\;}}\frac{1}{3}}}} \)

\( \Rightarrow \sqrt {\frac{4}{{{\rm{\;}}\frac{2}{3}}}} \)

⇒ √6

∴ The value is √6.