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

$\frac{1}{5}(1+2 \sqrt{3})$

7 sin²θ - cos²θ + 2sinθ = 2

{ sin²θ + cos²θ = 1 }

7 sin²θ - ( 1 - sin²θ )  + 2sinθ = 2

7 sin²θ + sin²θ  + 2sinθ - 3 = 0 

8 sin²θ + 2sinθ - 3 = 0

8 sin²θ + 6sinθ - 4sinθ - 3 = 0

2 sinθ (4sinθ + 3 ) - 1 ( 4sinθ + 3 ) = 0

(2sinθ - 1 ).( 4sinθ + 3 ) = 0

Either (2sinθ - 1 ) = 0  or ( 4sinθ + 3 ) = 0 

( 4sinθ + 3 ) = 0  is not possible.

So, (2sinθ - 1 ) = 0

sinθ = \(\frac{1}{2}\)

{ sin 30º = \(\frac{1}{2}\) }

Now,

\(\frac{sec2θ + cot2θ}{cosec2θ + tan2θ}\)

= \(\frac{sec60º + cot60º}{cosec60º + tan60º}\)

= \(\frac{2 + 1/√3}{√3/2 + √3}\)

= \(\frac{2√3 + 1}{3+ 2}\)

= \(\frac{2√3 + 1}{5}\)