Saved Bookmarks
| 1. |
In `Delta`ABC if `2c cos^2(A/2)+ 2a Cos^2(C/2)` - 3b = 0. Prove that a, b, c are in AP |
|
Answer» Given equation is , `2c cos^2(A/2)+2a cos^2(C/2) - 3b = 0` We know, `cos2x = 2cos^2x - 1 => 2cos^2x = 1+cos2x` `:. c(1+cosA)+a(1+cosC) - 3b = 0` `=>c(1+(b^2+c^2-a^2)/(2bc))+a(1+(a^2+b^2-c^2)/(2ab)) - 3b = 0` `=>c((2bc+b^2+c^2-a^2)/(2bc))+a((2ab+a^2+b^2-c^2)/(2ab)) - 3b = 0` `=>1/(2b)(2bc+b^2+c^2-a^2+2ab+a^2+b^2-c^2) - 3b =0` `=>1/(2b)(2b^2+2b(c+a)) - 3b = 0` `=>a+b+c - 3b = 0` `=>a+c = 2b` `=> b = (a+c)/2` So, `a,b and c` are in `A.P.` |
|