If $3 \sin ^2 \theta+4 \cos \theta-4=0,0^{\circ}<\theta<90^{\circ}$, then the value of $\left(cosec^2 \theta+\cot ^2 \theta\right)$ is
Answer & explanation
Correct answer: option 1
3 sin2θ + 4 cosθ - 4 = 0
3 ( 1 - cos2θ) + 4 cosθ - 4 = 0
3cos2θ - 4 cosθ + 1 = 0
on solving ,
cosθ = \(\frac{1}{3}\)
By using pythagoras theorem ,
P2 + B2 = H2
P2 + 12 = 32
P = 2√2
Now , ( cosec2θ + cot2θ)
= (\(\frac{3}{2√2}\))2 + (\(\frac{1}{2√2}\))2
= \(\frac{9}{8}\) + \(\frac{1}{8}\)
= \(\frac{5}{4}\)