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Question
If a cotθ = b, prove that `("a"sinθ - "b"cosθ)/("a"sinθ + "b"cosθ) = ("a"^2 - "b"^2)/("a"^2 + "b"^2)`
Solution
a cotθ = b
⇒ cotθ = `"b"/"a"`
⇒ tanθ = `(1)/cotθ = "a"/"b"`
To prove: `("a"sinθ - "b"cosθ)/("a"sinθ + "b"cosθ) = ("a"^2 - "b"^2)/("a"^2 + "b"^2)`
Consider `("a"sinθ - "b"cosθ)/("a"sinθ + "b"cosθ)`
Dividing the numerator and denominator by cosθ, we get
`("a"sinθ - "b"cosθ)/("a"sinθ + "b"cosθ)`
= `("a"sinθ/cosθ - "b")/("a"sinθ/cosθ + "b")`
= `("a"tanθ - "b")/("a"tanθ + "b")`
= `("a" xx "a"/"b" - "b")/("a" xx "a"/"b" + "b")`
= `(("a"^2 - "b"^2)/"b"^2)/(("a"^2 + "b"^2)/"b"^2`
= `("a"^2 - "b"^2)/("a"^2 - "b"^2)`.
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