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प्रश्न
In Δ ABC, if ∠C = 90°, then prove that sin (A - B) = `("a"^2 - "b"^2)/("a"^2 + "b"^2)`
उत्तर
In Δ ABC, if ∠C = 90°
∴ c2 = a2 + b2 .........(1)
By sine rule,
`"a"/"sin A" = "b"/"sin B" = "c"/"sin C"`
∴ `"a"/"sin A" = "b"/"sin B" = "c"/("sin" 90°)`
∴ `"a"/"sin A" = "b"/"sin B" = "c"` .....[∵ sin 90° = 1]
∴ sin A = `"a"/"c" and "sin B" = "b"/"c"` ....(2)
LHS = sin (A - B)
= sin A cos B - cos A sin B
`= "a"/"c" cos "B" - "b"/"c" cos "A"` ....[By (2)]
`= "a"/"c" (("c"^2 + "a"^2 - "b"^2)/"2ca") - "b"/"c"(("b"^2 + "c"^2 - "a"^2)/"2bc")`
`= ("c"^2 + "a"^2 - "b"^2)/"2c"^2 - ("b"^2 + "c"^2 - "a"^2)/"2c"^2`
`= ("c"^2 + "a"^2 - "b"^2 - "b"^2 - "c"^2 + "a"^2)/"2c"^2`
`= (2"a"^2 - 2"b"^2)/"2c"^2`
`= ("a"^2 - "b"^2)/"c"^2`
`= ("a"^2 - "b"^2)/("a"^2 + "b"^2)` ...[By (1)]
= RHS.
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