If $3 \sin ^2 \theta-\cos \theta-1=0,0^{\circ}<\theta<90^{\circ}$, then what is the value of $\cot \theta+{cosec} \theta$ ?
Answer & explanation
Correct answer: option 3
We are given that ,
3 sin²θ - cosθ - 1 = 0
{ using , sin²θ + cos²θ = 1 }
3 ( 1 - cos²θ ) - cosθ - 1 = 0
3 cos²θ + cosθ - 2 = 0
3 cos²θ + 3 cosθ - 2cosθ - 2 = 0
3 cosθ (cosθ + 1 ) - 2 (cosθ + 1 ) = 0
(3cosθ - 2 ) . (cosθ + 1 ) = 0
Either (3cosθ - 2 ) = 0 or (cosθ + 1 ) = 0
(cosθ + 1 ) = 0 is not possible
So, (3cosθ - 2 ) = 0
cosθ = \(\frac{2}{3}\)
{ we know, cosθ = \(\frac{B}{H}\) }
By using pythagoras theorem,
P² + B² = H²
P² + 2² = 3²
P = √5
Now,
cotθ+ cosecθ
= \(\frac{B}{P}\) + \(\frac{H}{P}\)
= \(\frac{2}{√5}\) + \(\frac{3}{√5}\)
= \(\frac{5}{√5}\)
= √5