If $5 \sin ^2 \theta-4 \cos \theta-4=0,0^{\circ}<\theta<90^{\circ}$, then the value of $(\cot \theta+{cosec} \theta)$ is:
Answer & explanation
Correct answer: option 1
We are given that :-
5sin²θ - 4cosθ - 4 = 0
{ using , sin²θ + cos²θ = 1 }
5( 1 - cos²θ) - 4cosθ - 4 = 0
5 cos²θ + 4cosθ - 1 = 0
5 cos²θ + 5cosθ - cosθ - 1 = 0
5 cosθ (cosθ + 1 ) - 1 ( cosθ + 1 ) = 0
Either (5 cosθ - 1 ) = 0 or ( cosθ + 1 ) = 0
(cosθ + 1 ) = 0 is not possible
So, (5 cosθ - 1 ) = 0
cosθ = \(\frac{1}{5}\)
By using pythagoras theorem,
P² + B² = H²
P² + 1² = 5²
P² = 24
P = 2√6
Now,
( cotθ + cosecθ )
= \(\frac{B}{P}\) + \(\frac{H}{P}\)
= \(\frac{1}{2√6}\) + \(\frac{5}{2√6}\)
= \(\frac{3}{√6}\)
= \(\frac{3}{√6}\) × \(\frac{√6}{√6}\)
= \(\frac{√6}{2}\)