In $\triangle ABC$, with usual notations prove that $b^2 = c^2 + a^2 - 2ca\cos B$ **OR** In $\triangle ABC$, with usual notations prove that $(a-b)^2\cos^2\left(\dfrac{C}{2}\right) + (a+b)^2\sin^2\left(\dfrac{C}{2}\right) = c^2$.

In triangle ABC, with usual notations prove that b – Class 12 Mathematics

Question 151 of 177

\triangle ABC

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2019Subjective· 4mImportance★★★★★

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Part 1: drop a perpendicular and apply Pythagoras. Part 2 (OR): expand the half-angle expression and use the cosine rule.

Part 1 — Prove $b^2=c^2+a^2-2ca\cos B$ :

In $\triangle ABC$ , draw $AD\perp BC$ with $D$ on line $BC$ .

In right $\triangle ADB$ : $AD = c\sin B$ , $BD = c\cos B$ .

Then $DC = BC-BD = a-c\cos B$ .

In right $\triangle ADC$ , by Pythagoras: $AC^2 = AD^2+DC^2$

$b^2 = (c\sin B)^2 + (a-c\cos B)^2 = c^2\sin^2B + a^2 - 2ac\cos B + c^2\cos^2B$

$= c^2(\sin^2B+\cos^2B) + a^2 - 2ac\cos B = c^2+a^2-2ac\cos B$

i.e. $b^2 = c^2+a^2-2ca\cos B$ . Hence proved.

Part 2 (OR) — Prove $(a-b)^2\cos^2\left(\dfrac C2\right)+(a+b)^2\sin^2\left(\dfrac C2\right)=c^2$ :

$(a-b)^2\cos^2\left(\dfrac C2\right)+(a+b)^2\sin^2\left(\dfrac C2\right)$

$= (a^2-2ab+b^2)\cos^2\left(\dfrac C2\right) + (a^2+2ab+b^2)\sin^2\left(\dfrac C2\right)$

…

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

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

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

…

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

…

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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 $c^2+a^2-b^2=ac$ : …

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CBSE 2017Set ANNUAL2 marks
$\triangle ABC$
›Reveal solutionSolution

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

By the cosine rule:

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

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

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

…

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Q.In $\triangle ABC$ , with usual notations prove that $b^2 = c^2 + a^2 - 2ca\cos B$ OR In $\triangle ABC$ , with usual notations prove that $(a-b)^2\cos^2\left(\dfrac{C}{2}\right) + (a+b)^2\sin^2\left(\dfrac{C}{2}\right) = c^2$ .

Concept understanding — Cosine Rule

Cosine Rule

Where it comes from

How it is used

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