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In any Δ ABC, prove the following: a2 sin (B - C) = (b2 - c2) sin A) - Mathematics and Statistics

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प्रश्न

In any Δ ABC, prove the following:

a2 sin (B - C) = (b2 - c2) sin A.

योग

उत्तर

By sine rule,

`"a"/"sin A" = "b"/"sin B" = "c"/"sin C" = "k"`

∴ a = k sin A, b = k sin B, c = k sin C

RHS = (b2 - c2) sin A

= (k2 sin2B - k2 sin2C)sin A

= k2 (sin2B - sin2C) sin A

= k2 (sin B + sin C)(sin B - sin C) sin A

= `"k"^ 2 xx  2 "sin" (("B + C")/2). cos(("B - C")/2) xx 2 cos (("B + C")/2).sin (("B - C")/2) xx sin "A"`

= `"k"^ 2 xx  2 "sin" (("B + C")/2). cos(("B + C")/2) xx 2 sin (("B - C")/2). cos (("B - C")/2) xx sin "A"`

= k2 x sin(B + C) x sin (B - C) x sin A

= k. sin (π - A). sin (B - C). sin A    ....[∵ A + B + C = π]

= k2. sin A. sin (B - C). sin A

= (k sin A)2. sin(B - C)

= a2 sin (B - C)

= LHS

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अध्याय 3: Trigonometric Functions - Miscellaneous exercise 3 [पृष्ठ १०९]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 11.3 | पृष्ठ १०९

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