Question

a(cosC - cosB)=2(b-c)cos^2.A/2​

Prove that sum


please guys I need this answer

Answer 1

✪ᴀɴsᴡᴇʀ✪

$L. H. S=$

$a(cosC - cosB)$

$a\left(\frac{a^2 +b^2 - c^2 } {2ab} - \frac{c^2 +a^2 - b^2 }{2ca} \right)$

$\frac{a}{2abc} \left[c(a^2 +b^2 - c^2) - b(c^2 +a^2 - b^2) \right]$

$\frac{1}{2bc} \left[ca^2 + b^2c - c^3 - bc^2 - a^2b + b^3\right]$

$\frac{1}{2bc} \left[(b^3 - c^3) + (b^2c-bc^2) -(a^2b - ca^2) \right]$

$\frac{1}{2bc} \left[(b - c)[b^2 + bc + c^2] + bc(b - c) -a^2(b - c) \right]$

$\frac{1}{2bc}(b - c)[b^2 + bc + c^2 + bc -a^2]$

$\frac{1}{2bc}(b - c)[(b^2 + c^2 -a^2) + 2bc ]$

$= (b - c)[\frac{(b^2 + c^2 -a^2)}{2bc}+ 1 ]$

$= (b - c)[1 + cosA]$

$= (b - c)[2cos^2(\frac{A}{2 })]$

$= 2(b - c)cos^2(\frac{A}{2 })$

$= R. H. S$