This is the basic and exam oriented question from trigonometric identities in which we have given that (2sinθ+3cosθ)=2, and we need to prove that (3sinθ-2cosθ)=±3
Kindly give me a detailed solution of this question
RS Aggarwal, Class 10, chapter 13B, question no 10
Given 2sinθ + 3cosθ = 2
Now
(3sinθ−2cosθ)²
=(9sin²θ−2⋅3sinθ⋅2cosθ+4cos²θ
=9−9cos²θ−2⋅3cosθ⋅2sinθ+4−4sin²θ
=13−[(3cosθ)²+2⋅3cosθ⋅2sinθ+(2sinθ)²]
=13−(2sinθ+3cosθ)²
=13−2² = 9
So
(3sinθ−2cosθ)² = 9
⇒3sinθ−2cosθ = ±√9
= ±3