If $2 \sin \theta+15 \cos ^2 \theta=7,0^{\circ}<\theta<90^{\circ}$, then $\tan \theta+\cos \theta+\sec \theta=$

$3 \frac{3}{5}$

2 sinθ + 15 cos²θ  = 7

{ sin²θ + cos²θ = 1 }

2 sinθ + 15 ( 1 - sin²θ)  = 7

2 sinθ + 15 ( 1 - sin²θ)  = 7

15 sin²θ  - 2 sinθ -  8 = 0 

15 sin²θ  - 12 sinθ + 10 sinθ -  8 = 0

3 sinθ (5 sinθ - 4) + 2 ( 5 sinθ - 4) = 0

(3 sinθ + 2) . ( 5 sinθ - 4) = 0

Either (3 sinθ + 2)= 0 or  ( 5 sinθ - 4) = 0 

(3 sinθ + 2)= 0 is not possible.

So, ( 5 sinθ - 4) = 0 

sinθ = \(\frac{4}{5}\)

{ sinθ = \(\frac{P}{H}\) }

By using pythagoras theorem,

P² + B² = H²

4² + B² = 5²

B = 3

Now,

tanθ + cosθ + secθ

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

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

= \(\frac{20 + 9 + 25}{15}\)

= 3\(\frac{ 3}{5}\)