In $\triangle ABC$ prove that $ac\cos B - bc\cos A = a^2 - b^2$.

In triangle ABC prove that accos B - bccos A = a² – Class 12 Mathematics

Miscellaneous Exercise 3 · Q106

\triangle ABC

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

60% · 106/177 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

By the Cosine Rule, $\cos B=\dfrac{a^2+c^2-b^2}{2ac}$ and $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$ . So $ac\cos B=\dfrac{a^2+c^2-b^2}{2}$ and $bc\cos A=\dfrac{b^2+c^2-a^2}{2}$ . Subtracting: $ac\cos B-bc\cos A=\dfrac{(a^2+c^2-b^2)-( …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL2 marks
$\triangle ABC$
›Reveal solutionSolution

Substitute $\cos C$ and $\cos B$ from the cosine rule and simplify.

By the cosine rule:

$\cos C=\frac{a^2+b^2-c^2}{2ab}, \qquad \cos B=\frac{a^2+c^2-b^2}{2ac}$

LHS $=a(b\cos C-c\cos B) = ab\cos C - ac\cos B$

$ab\cos C = ab\cdot\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2-c^2}{2}$

$ac\cos B = ac\cdot\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+c^2-b^2}{2}$

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL2 marksMCQ
$\triangle ABC$
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{2}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

Compare with the cosine rule $b^2=a^2+c^2-2ac\cos B$ .

By the cosine rule, $b^2=a^2+c^2-2ac\cos B \Rightarrow a^2+c^2-b^2=2ac\cos B$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL2 marksMCQ
$\triangle ABC$
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{2}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

Compare with the Cosine Rule formula for $\cos B$ .

By the Cosine Rule: $\cos B = \dfrac{a^2+c^2-b^2}{2ac}$

Given $c^2+a^2-b^2=ac$ : …

Start your 14-day free trial to see the full solution →
CBSE 2017Set ANNUAL2 marks
$\triangle ABC$
›Reveal solutionSolution

Substitute $\cos B$ and $\cos C$ from the cosine rule and simplify.

By the cosine rule:

$\cos C = \dfrac{a^2+b^2-c^2}{2ab}, \qquad \cos B = \dfrac{a^2+c^2-b^2}{2ac}$

$a(b\cos C - c\cos B) = a\left[b\cdot\dfrac{a^2+b^2-c^2}{2ab} - c\cdot\dfrac{a^2+c^2-b^2}{2ac}\right]$

$= a\left[\dfrac{a^2+b^2-c^2}{2a} - \dfrac{a^2+c^2-b^2}{2a}\right] = \dfrac{1}{2}\Big[(a^2+b^2-c^2)-(a^2+c^2-b^2)\Big]$

…

Start your 14-day free trial to see the full solution →

←Previous questionQ105 — In triangle ABC prove that a²sin(B-C) = (b²-c²)sin A.→Next questionQ107 — In triangle ABC prove that (cos A)/(a) + (cos B)/(b) + (cos C)/(c) = (…

Q.In $\triangle ABC$ prove that $ac\cos B - bc\cos A = a^2 - b^2$ .

Concept understanding — Cosine Rule

Cosine Rule

Where it comes from

How it is used

Q.In △ABC\triangle ABC△ABC prove that accos⁡B−bccos⁡A=a2−b2ac\cos B - bc\cos A = a^2 - b^2accosB−bccosA=a2−b2.

🎓Unlock everything free for 14 days

Q.In $\triangle ABC$ prove that $ac\cos B - bc\cos A = a^2 - b^2$ .

Unlock everything free for 14 days