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प्रश्न
If 4 sinθ = 3 cosθ, find `(6sinθ - 2cosθ )/(6sinθ + 2cosθ )`
उत्तर
4 sinθ = 3 cosθ
⇒ `(sin θ)/(cos θ) = (3)/(4)`
⇒ tan θ = `(sin θ)/(cos θ) = (3)/(4)`
⇒ cot θ = `(1)/(tanθ) = (4)/(3)`
`(6sinθ - 2cosθ )/(6sinθ + 2cosθ )`
= `(6 xx sinθ/cosθ - 2 xx cosθ/cosθ)/(6 xx sinθ/cosθ + 2 xx cosθ/cosθ)`
= `(6tan θ - 2)/(6tan θ + 2)`
= `(6 xx 3/4 - 2)/(6 xx 3/4 + 2)`
= `(9/2 - 2)/(9/2 + 2)`
= `((9 - 4)/(2))/((9 + 4)/(2)`
= `(5)/(13)`.
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