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Question Number 20019 by Tinkutara last updated on 20/Aug/17

$$\mathrm{In}\mathrm{any}\mathrm{triangle}ABC,\mathrm{prove}\mathrm{that}$$$$a(\cos C-\cos B)=2(b-c){\cos }^2\frac{A}{2}$$

Answered by ajfour last updated on 20/Aug/17

$$l.h.s.=a[\frac{a^2+b^2-c^2}{2ab}-\frac{c^2+a^2-b^2}{2ac}]$$$$=\frac{c[a^2+(b^2-c^2)]+b[(b^2-c^2)-a^2]}{2bc}$$$$=\frac{(b^2-c^2)(b+c)-a^2(b-c)}{2bc}$$$$=\frac{(b-c)[{(b+c)}^2-a^2]}{2bc}$$$$=(b-c)\frac{(2bc+b^2+c^2-a^2)}{2bc}$$$$=(b-c)(1+\frac{b^2+c^2-a^2}{2bc})$$$$=(b-c)(1+\cos A)$$$$=2(b-c)\cos {}^2\frac{A}{2}=r.h.s.$$

Commented by Tinkutara last updated on 20/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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