In $\triangle ABC$ prove that $\dfrac{1-\cos 2A}{a^2} = \dfrac{1-\cos 2B}{b^2}$.

In triangle ABC prove that (1-cos 2A)/(a²) = (1-cos 2B)/(b²).

Miscellaneous Exercise 3 · Q108

\triangle ABC

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$1-\cos2A=2\sin^2A$ and $1-\cos2B=2\sin^2B$ . So $\dfrac{1-\cos2A}{a^2}=\dfrac{2\sin^2A}{a^2}$ and $\dfrac{1-\cos2B}{b^2}=\dfrac{2\sin^2B}{b^2}$ . By the Sine Rule, $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}\implies\dfrac{\sin^2A}{a^2}=\dfrac{\sin^2B}{b^2}$ , so $\dfrac{2\sin^2A …

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Q.In $\triangle ABC$ prove that $\dfrac{1-\cos 2A}{a^2} = \dfrac{1-\cos 2B}{b^2}$ .

Concept understanding — Solving a Triangle - Sine Rule

Q.In △ABC\triangle ABC△ABC prove that 1−cos⁡2Aa2=1−cos⁡2Bb2\dfrac{1-\cos 2A}{a^2} = \dfrac{1-\cos 2B}{b^2}a21−cos2A​=b21−cos2B​.

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Q.In $\triangle ABC$ prove that $\dfrac{1-\cos 2A}{a^2} = \dfrac{1-\cos 2B}{b^2}$ .

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