RRB NTPC · Question 8
2cos(π/2 – θ) + 3sin(π/2 + θ) – (3sinθ + 2cosθ) = ? A. cosθ – sin θ B. sinθ – cosθ C. sinθ + cosθ D. cotθ – tanθ
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Answer: (D) A
Explanation
cos(π/2 − θ) = sin θ and sin(π/2 + θ) = cos θ. So the expression = 2sin θ + 3cos θ − 3sin θ − 2cos θ = cos θ − sin θ.