If 5 sinθ + 3 cosθ = 4, find the value of 3 sinθ - 5 cosθ.
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hi friend,
5 sinθ + 3 cosθ = 4
→squaring on both sides ,we get
25sin²θ+9cos²θ+30sinθcosθ =16-----(1)
now let us assume that
3 sinθ - 5 cosθ.=x
squaring on both sides, we get
→9sin²θ+25cos²θ-30sinθcosθ=x²-----(2)
adding (1) and (2)
→34sin²θ+34cos²θ=16+x²
→34(sin²θ+cos²θ)=16+x²
→34-16=x²
→x=±√18
so, 3 sinθ - 5 cosθ. =±√18
I hope this will help u :)
5 sinθ + 3 cosθ = 4
→squaring on both sides ,we get
25sin²θ+9cos²θ+30sinθcosθ =16-----(1)
now let us assume that
3 sinθ - 5 cosθ.=x
squaring on both sides, we get
→9sin²θ+25cos²θ-30sinθcosθ=x²-----(2)
adding (1) and (2)
→34sin²θ+34cos²θ=16+x²
→34(sin²θ+cos²θ)=16+x²
→34-16=x²
→x=±√18
so, 3 sinθ - 5 cosθ. =±√18
I hope this will help u :)
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Prove that following identities using A=60.
(i) Sin^2 A + Cos^2 A=1 (ii) Sec^2 A - Tan^2 A=1 (iii) Cosec^2 A - Cot^2 A=1
(i) Sin^2 A + Cos^2 A=1 (ii) Sec^2 A - Tan^2 A=1 (iii) Cosec^2 A - Cot^2 A=1
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