Q.4 In triangle ABC , prove that a(bcos C - ccos B) = b² - c² .

Answer: Proved using the cosine rule. By the cosine rule, cos C = a² + b² - c²/2ab and cos B = a² + c² - b²/2ac . L.H.S. = abcos C - accos B = a² + b² - c²/2 - a² + c² - b²/2 = 2b² - 2c²/2 = b² - c² = R.H.S.

In \(\triangle ABC\), prove that \(a(b\cos C - c\cos B) = b^2 - c^2\).

Q.4 In \(\triangle ABC\), prove that \(a(b\cos C - c\cos B) = b^2 - c^2\).

Doubtly Asked 28 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: Proved using the cosine rule.

By the cosine rule, \(\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}\) and \(\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}\).

L.H.S. \(= ab\cos C - ac\cos B = \dfrac{a^2 + b^2 - c^2}{2} - \dfrac{a^2 + c^2 - b^2}{2} = \dfrac{2b^2 - 2c^2}{2} = b^2 - c^2 =\) R.H.S.

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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In \(\triangle ABC\), prove that \(a(b\cos C - c\cos B) = b^2 - c^2\).

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