If $5 \sin ^2 \theta+14 \cos \theta=13,0^{\circ}<\theta<90^{\circ}$, then what is the value of $\frac{\sec \theta+\cot \theta}{{cosec} \theta+\tan \theta}$ ?
Answer & explanation
Correct answer: option 3
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,
\(\frac{secθ +cotθ}{cosecθ + tanθ}\)
= \(\frac{B/H +B/P}{H/P + P/B}\)
= \(\frac{4/5 +4/3}{5/3 + 3/4}\)
= \(\frac{31/12}{ 29/12 }\)
= \(\frac{31}{ 29 }\)