<em>θ</em> is given to be in the fourth quadrant (270° < <em>θ</em> < 360°) for which sin(<em>θ</em>) < 0 and cos(<em>θ</em>) > 0. This means
cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1 ==> sin(<em>θ</em>) = -√[1 - cos²(<em>θ</em>)] = -3/5
Now recall the double angle identity for sine:
sin(2<em>θ</em>) = 2 sin(<em>θ</em>) cos(<em>θ</em>)
==> sin(2<em>θ</em>) = 2 (-3/5) (4/5) = -24/25
Answer is: B
Step-by-step explanation: correct answer is B
b
2
−a
2
Given, acotθ+bcscθ=p
bcotθ+acscθ=q
⇒p
2
−q
2
=(p−q)(p+q)
=[acotθ+bcscθ−bcotθ−acscθ][acotθ+bcscθ+bcotθ+acscθ]
=[cotθ(a−b)−cscθ(a−b)][cotθ(a+b)+cscθ(a+b)]
=(a−b)(cotθ−cscθ)(a+b)(cotθ+cscθ)
=(a
2
−b
2
)(cot
2
θ−csc
2
θ)
=(−1)(a
2
−b
2
)[∵csc
2
θ−cot
2
θ=1]
=(b
2
−a
2
)
