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In Δ ABC, prove that a2 (cos2 B - cos2 C) + b2 (cos2 C - cos2 A) + c2 (cos2 A - cos2 B) = 0.

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Question

In Δ ABC, prove that a2 (cos2 B - cos2 C) + b2 (cos2 C - cos2 A) + c2 (cos2 A - cos2 B) = 0.

Sum
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Solution

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 

LHS = `"a"^2 (cos^2"B" - cos^2"C") + "b"^2(cos^2 "C" - cos^2 "A") + "c"^2 (cos^2"A" - cos^2"B")`

`= "k"^2 sin^2"A" [(1 - sin^2"B") - (1 - sin^2"C")] + "k"^2 sin^2"B" [(1 - sin^2"C") - (1 - sin^2"A")] + "k"^2 sin^2"C" [(1 - sin^2"A") - (1 - sin^2"B")]`

`= "k"^2 sin^2"A" (sin^2"C" - sin^2"B") + "k"^2sin^2"B"(sin^2"A" - sin^2"C") + "k"^2 sin^2"C" (sin^2"B" - sin^2"A")`

`= "k"^2 (sin^2"A" sin^2"C" - sin^2"A" sin^2"B" + sin^2"A" sin^2"B" - sin^2"B" sin^2"C" + sin^2"B" sin^2"C" - sin^2"A" sin^2"C")`

`= "k"^2 (0)`

= 0

= RHS.

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Chapter 3: Trigonometric Functions - Miscellaneous exercise 3 [Page 110]

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Balbharati Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 16 | Page 110

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