If $2 \sin \theta+15 \cos ^2 \theta=7,0^{\circ}<\theta<90^{\circ}$, then $\tan \theta+\cos \theta+\sec \theta=$
Answer & explanation
Correct answer: option 1
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\(\frac{ 3}{5}\)