With usual notations show that $(c^2-a^2+b^2) an A = (a^2-b^2+c^2) an B = (b^2-c^2+a^2) an C$.

Question

With usual notations show that $(c^2-a^2+b^2)	an A = (a^2-b^2+c^2)	an B = (b^2-c^2+a^2)	an C$.

Accepted Answer

Each bracket is the Cosine Rule's numerator for one specific angle: $b^2+c^2-a^2=2bc\cos A$ (i.e. $c^2-a^2+b^2=2bc\cos A$ ), $a^2+c^2-b^2=2ac\cos B$ (i.e. $a^2-b^2+c^2=2ac\cos B$ ), and $a^2+b^2-c^2=2ab\cos C$ (i.e. $b^2-c^2+a^2=2ab\cos C$ ). Multiplying each by $\tan$ of its matching angle converts $\cos$ to $\sin$ : $(c^2-a^2+b^2)\tan A=2bc\cos A\tan A=2bc\sin A=2bc\cdot\dfrac{a}{2R}=\dfrac{abc}{R}$ (Sine Rule, $\sin A=a/2R$ ). By the same argument, $(a^2-b^2+c^2)\tan B=2ac\sin B=\dfrac{abc}{R}$ and $(b^2-c^2+a^2)\tan C=2ab\sin C=\dfrac{abc} …

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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.With usual notations show that $(c^2-a^2+b^2)\tan A = (a^2-b^2+c^2)\tan B = (b^2-c^2+a^2)\tan C$ .

Concept understanding — Cosine Rule

Cosine Rule

Where it comes from

How it is used

Q.With usual notations show that (c2−a2+b2)tan⁡A=(a2−b2+c2)tan⁡B=(b2−c2+a2)tan⁡C(c^2-a^2+b^2)\tan A = (a^2-b^2+c^2)\tan B = (b^2-c^2+a^2)\tan C(c2−a2+b2)tanA=(a2−b2+c2)tanB=(b2−c2+a2)tanC.

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Q.With usual notations show that $(c^2-a^2+b^2)\tan A = (a^2-b^2+c^2)\tan B = (b^2-c^2+a^2)\tan C$ .

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