Trigonometric Functions — Maths STD 12 Science — Question

$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}=k$

$\therefore a=k \sin \mathrm{A}, b=k \sin \mathrm{B}, c=k \sin \mathrm{C}$

$\mathrm{RHS}=\left(\frac{b-c}{a}\right) \cos \frac{\mathrm{A}}{2}$

$=\left(\frac{k \sin \mathrm{B}-k \sin \mathrm{C}}{k \sin \mathrm{A}}\right) \cos \frac{\mathrm{A}}{2}$$=\left(\frac{k \sin \mathrm{B}-k \sin \mathrm{C}}{k \sin \mathrm{A}}\right) \cos \frac{\mathrm{A}}{2}$

$=\left(\frac{\sin B-\sin C}{\sin A}\right) \cos \frac{A}{2}$

$=\frac{2 \cos \left(\frac{B+C}{2}\right) \cdot \sin \left(\frac{B-C}{2}\right)}{2 \sin \frac{A}{2} \cdot \cos \frac{A}{2}} \cdot \cos \frac{A}{2}$

$=\frac{\cos \left(\frac{B+C}{2}\right) \cdot \sin \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}$

$=\frac{\cos \left(\frac{\pi}{2}-\frac{A}{2}\right) \cdot \sin \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}} \ldots[\because A+B+C=\pi]$

$=\frac{\sin \frac{A}{2} \cdot \sin \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}$

$=\sin \left(\frac{B-C}{2}\right)=$ LHS