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Question
If 3 cot θ = 2 then prove that `((4 sin θ - 3 cos θ))/((2 sin θ + 6 cos θ)) = 1/3`.
Theorem
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Solution
Given: 3 cot θ = 2
To Prove: `((4 sin θ - 3 cos θ))/((2 sin θ + 6 cos θ)) = 1/3`
Proof [Step-wise]:
1. From 3 cot θ = 2, divide both sides by 3: `cot θ = 2/3`.
2. By definition `cot θ = (cos θ)/(sin θ)`, so `cos θ = (2/3) sin θ`.
3. Substitute `cos θ = (2/3) sin θ` into the numerator:
4 sin θ – 3 cos θ
= `4 sin θ - 3[(2/3) sin θ]`
= 4 sin θ – 2 sin θ
= 2 sin θ
4. Substitute cos θ into the denominator:
2 sin θ + 6 cos θ
= `2 sin θ + 6[(2/3) sin θ]`
= 2 sin θ + 4 sin θ
= 6 sin θ
5. Form the quotient:
`(4 sin θ - 3 cos θ)/(2 sin θ + 6 cos θ)`
= `(2 sin θ)/(6 sin θ)`
= `2/6`
= `1/3`
Hence, `(4 sin θ - 3 cos θ)/(2 sin θ + 6 cos θ) = 1/3`, as required.
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