If $3 \sec ^2 \theta+\tan \theta=7,0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{{cosec} 2 \theta+\cos \theta}{\sin 2 \theta+\cot \theta}$ is:

$\frac{2+\sqrt{2}}{4}$

3 sec²θ + tanθ = 7

{ we know,  sec²θ - tan²θ = 1 }

3 ( 1 + tan²θ ) + tanθ = 7

3 tan²θ  + tanθ - 4 = 0

3 tan²θ  +4 tanθ - 3tanθ  - 4 = 0

tanθ ( 3tanθ + 4 ) - 1 ( 3tanθ + 4 ) = 0

( tanθ - 1 ). ( 3tanθ + 4 ) = 0

Either ( tanθ - 1 ) = 0 Or  ( 3tanθ + 4 ) = 0 

( 3tanθ + 4 ) = 0  is not possible.

So, ( tanθ - 1 ) = 0

  tanθ = 1

{ tan45º = 1 }

Now,

\(\frac{ cosec 2θ + cosθ}{ sin2θ + cotθ}\)

= \(\frac{ cosec 90º + cos45º}{ sin90º + cot45º}\)

= \(\frac{ 1 +1/√2}{ 1 +1 }\)

= \(\frac{ √2 +1}{ 2√2 }\)

= \(\frac{ 2 +√2}{ 4 }\)