If $\cot \theta+\tan \theta=2 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\tan 2 \theta-\sec \theta}{\cot 2 \theta+{cosec} \theta}$.

$\frac{2 \sqrt{3}-1}{11}$

cot θ + tan θ = 2 sec θ

\(\frac{cosθ}{sinθ}\) + \(\frac{sinθ}{cosθ}\) = 2 . \(\frac{1}{cosθ}\)

\(\frac{cos²θ + sin²θ}{sinθcosθ}\) = 2 . \(\frac{1}{cosθ}\)

{ cos²θ + sin²θ = 1 }

\(\frac{1}{sinθ}\) = 2

sinθ = \(\frac{1}{2}\)

{ we know, sin 30º = \(\frac{1}{2}\) }

So, θ = 30º

Now,

\(\frac{tan2θ - secθ}{cot2θ + cosecθ}\)

= \(\frac{tan60º - sec30º}{cot60º + cosec30º}\)

= \(\frac{ √3  - 2/√3 }{1/√3 + 2}\)

= \(\frac{ 1 }{  2√3 + 1 }\)

= \(\frac{ 1 }{  2√3 + 1 }\) × \(\frac{  2√3 - 1 }{  2√3 - 1 }\) 

= \(\frac{  2√3 - 1 }{  11}\)