If $5 \sin ^2 \theta+14 \cos \theta=13,0^{\circ}<\theta<90^{\circ}$, then what is the value of $\frac{\sec \theta+\cot \theta}{{cosec} \theta+\tan \theta}$ ?

$\frac{31}{29}$

5 sin²θ + 14 cosθ = 13

{ using , sin²θ + cos²θ = 1 }

5 ( 1 - cos²θ ) + 14 cosθ = 13

5 cos²θ - 14 cosθ + 8 = 0 

5 cos²θ - 10 cosθ - 4 cosθ + 8 = 0

5 cosθ (cosθ - 2) - 4 ( cosθ - 2 ) = 0

(5 cosθ - 4) . ( cosθ - 2 ) = 0

EIther (5 cosθ - 4) = 0 Or  ( cosθ - 2 ) = 0 

( cosθ - 2 ) = 0  is not possible.

So, 5 cosθ - 4 = 0

cosθ = \(\frac{4}{5}\)

{ cosθ = \(\frac{B}{H}\) }

By using pythagoras theorem,

P² + B² = H²

P² + 4² = 5²

P = 3

Now,

\(\frac{secθ +cotθ}{cosecθ + tanθ}\)

= \(\frac{B/H +B/P}{H/P + P/B}\)

= \(\frac{4/5 +4/3}{5/3 + 3/4}\)

= \(\frac{31/12}{ 29/12  }\)

= \(\frac{31}{ 29  }\)