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If 3 cot θ = 2 then prove that ((4 sin θ – 3 cos θ))/((2 sin θ + 6 cos θ)) = 1/3.

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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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Chapter 10: Trignometric Ratios - EXERCISE 10 [Page 547]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 10 Trignometric Ratios
EXERCISE 10 | Q 15. | Page 547
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