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प्रश्न
If sin θ = `"a"/sqrt("a"^2 + "b"^2)`, then show that b sin θ = a cos θ
उत्तर
sin θ = `"a"/sqrt("a"^2 + "b"^2)`
In the triangle ΔABC
BC2 = AC2 – AB2
= `(sqrt("a"^2 + "b"^2))^2 - "a"^2`
= a2 + b2 − a2 = b2
BC = `sqrt("b"^2)` = b
cos θ = `"b"/sqrt("a"^2 + "b"^2)`
L.H.S = b sin θ
= `"b" "a"/sqrt("a"^2 + "b"^2)`
= `"ab"/sqrt("a"^2 + "b"^2)`
R.H.S = a cos θ
= `"a" "b"/sqrt("a"^2 + "b"^2)`
= `"ab"/sqrt("a"^2 + "b"^2)`
L. H. S = R. H. S
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