Q.In $\triangle ABC$, prove that $\sin\left(\dfrac{B-C}{2}\right) = \left(\dfrac{b-c}{a}\right)\cos\left(\dfrac{A}{2}\right)$ OR Show that $\sin^{-1}\left(\dfrac{5}{13}\right) + \cos^{-1}\left(\dfrac{3}{5}\right) = \tan^{-1}\left(\dfrac{63}{16}\right)$

Part 1: use the sine rule + sum-to-product. Part 2 (OR): compute $\tan(\alpha+\beta)$ and check the range.

Part 1 — Prove $\sin\left(\dfrac{B-C}{2}\right) = \left(\dfrac{b-c}{a}\right)\cos\left(\dfrac A2\right)$:

By the sine rule, $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$, so $b=k\sin B$, $c=k\sin C$.

$\dfrac{b-c}{a} = \dfrac{k\sin B - k\sin C}{k\sin A} = \dfrac{\sin B-\sin C}{\sin A}$

Using sum-to-product: $\sin B-\sin C = 2\cos\left(\dfrac{B+C}{2}\right)\sin\left(\dfrac{B-C}{2}\right)$, and $\sin A = 2\sin\left(\dfrac A2\right)\cos\left(\dfrac A2\right)$.

Since $A+B+C=\pi$, $\dfrac{B+C}{2} = \dfrac\pi2-\dfrac A2$, so $\cos\left(\dfrac{B+C}{2}\right)=\sin\left(\dfrac A2\right)$.

$\dfrac{b-c}{a} = \dfrac{2\sin\left(\frac A2\right)\sin\left(\frac{B-C}{2}\right)}{2\sin\left(\frac A2\right)\cos\left(\frac A2\right)} = \dfrac{\sin\left(\frac{B-C}{2}\right)}{\cos\left(\frac A2\right)}$

$\implies \sin\left(\dfrac{B-C}{2}\right) = \left(\dfrac{b-c}{a}\right)\cos\left(\dfrac A2\right)$

Hence proved.

Part 2 (OR) — Show $\sin^{-1}\left(\dfrac{5}{13}\right)+\cos^{-1}\left(\dfrac35\right) = \tan^{-1}\left(\dfrac{63}{16}\right)$:

Let $\alpha=\sin^{-1}\left(\dfrac{5}{13}\right)$: $\sin\alpha=\dfrac{5}{13}$, $\cos\alpha=\dfrac{12}{13}$ (acute), so $\tan\alpha=\dfrac{5}{12}$.

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