If 3sinθ+5cosθ=5, prove that 5 sinθ-3cosθ=±3.
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3 sinθ + 5 cosθ = 5
Squaring both sides ,
( 3 sinθ + 5 cosθ )² = 5²
9 sin²θ + 25 cos²θ + 30 sinθ.cos θ = 25
9( 1 - cos²θ ) + 25( 1 -sin²θ ) + 30 sinθ.cosθ =25
Using identity , [ sin²θ + cos²θ =1 ]
9 - 9 cos²θ + 25 - 25 sin²θ +30 sinθ.cosθ =25
- 9 cos²θ - 25 sin²θ + 30 sinθcosθ = - 9
25 sin²θ + 9 cos²θ - 30 sinθ.cosθ = 9
( 5.sinθ - 3.cosθ )² = 9
[ 5 sinθ - 3 cosθ ) = ±3 ]
Squaring both sides ,
( 3 sinθ + 5 cosθ )² = 5²
9 sin²θ + 25 cos²θ + 30 sinθ.cos θ = 25
9( 1 - cos²θ ) + 25( 1 -sin²θ ) + 30 sinθ.cosθ =25
Using identity , [ sin²θ + cos²θ =1 ]
9 - 9 cos²θ + 25 - 25 sin²θ +30 sinθ.cosθ =25
- 9 cos²θ - 25 sin²θ + 30 sinθcosθ = - 9
25 sin²θ + 9 cos²θ - 30 sinθ.cosθ = 9
( 5.sinθ - 3.cosθ )² = 9
[ 5 sinθ - 3 cosθ ) = ±3 ]
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