Question

prove:1+3cosa-4cos^3a/1-cosa=(1+2cosa)^2

Answer 1

Prove that:  $\frac{1+3cos(a)-4cos^3(a)}{1-cos(a)}= [1+2cos(a)]^2$

First we will take the right side and simplify....

$[1+2cos(a)]^2\\ \\ =[1+2cos(a)]*[1+2cos(a)]\\ \\ =1+2cos(a)+2cos(a)+4cos^2(a)\\ \\ =1+4cos(a)+4cos^2(a)\\ \\ =\frac{1+4cos(a)+4cos^2(a)}{1}$

Now multiplying both numerator and denominator by $[1-cos(a)]$, we will get....

$\frac{[1+4cos(a)+4cos^2(a)][1-cos(a)]}{1-cos(a)}\\ \\ =\frac{1+4cos(a)+4cos^2(a)-cos(a)-4cos^2(a)-4cos^3(a)}{1-cos(a)}$

$=\frac{1+3cos(a)-4cos^3(a)}{1-cos(a)}$ (This is the given left side)

So, the given equation is proved.