यदि 2cos2θ – 5cosθ + 2 = 0 जहाँ, 0° < θ < 90° हैI तो \(\sqrt {\frac{{1{\rm{\;}} + {\rm{\;ta}}{{\rm{n}}^2}{\rm{\theta }}}}{{1{\rm{\;}} - {\rm{\;co}}{{\rm{t}}^2}{\rm{\theta }}}}} \) का मान ज्ञात कीजिये?

1
√3/2
2
√6
3
√2/3
4
1

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