If sin θ(2sin θ + 3) = 2, 0° < θ < 90°, then what is the value of (sec2 θ +cot2 θ - cos2 θ) ?
Answer & explanation
Correct answer: option 4
sin θ(2sin θ + 3) = 2
2sin² θ + 3sin θ - 2 = 0
2sin² θ + 4sin θ - sin θ - 2 = 0
2sin θ ( sin θ + 2 ) - 1 ( sin θ + 2 ) = 0
(sin θ + 2 ) . (2sin θ - 1 ) = 0
sin θ + 2 = 0 is not possible.
So, 2sin θ - 1
sin θ = \(\frac{1}{2}\)
{ we know, sin30º = \(\frac{1}{2}\) }
so, θ = 30º
Now,
( sec2 θ +cot2 θ - cos2 θ )
= ( sec2 30º+cot2 30º - cos2 30º )
= \(\frac{4}{3}\) + 3 - \(\frac{3}{4}\)
= \(\frac{16 + 36 - 9 }{12}\)
= \(\frac{43 }{12}\)