if tanA=2x(x+1)/2x+1 then find the value of sinA and cosA

tanA = 2x(x+1) / 2x+1 ---- given 
sec^2 A = 1+ tan^2 A --- identity. 

sec^2 A = 1 + [4x^2(x+1)^2] / (2x+1)^2 = [(2x+1)^2 + 4x^2(x+1)^2] / (2x+1)^2 
= [4x^4 + 8x^3 + 8x^2 + 4x +1 ] / (2x+1)^2 
= (2x^2 + 2x + 1)^2 / (2x+1)^2 This gives --secA = (2x^2 + 2x + 1) / (2x + 1) 
So cosA = (2x + 1) / (2x^2 + 2x + 1)

Now sin^2 A = (1 - cos^2 A ) = 1 - (2x+1)^2 / (2x^2 + 2x + 1)^2 
= [(2x^2 + 2x + 1)^2 - (2x+1)^2] / (2x^2 + 2x + 1)^2 
= (2x^2)*(2x^2 + 4x + 2 ) / (2x^2 + 2x + 1)^2 ---- using [a^2-b^2 =(a+b)*(a-b) ] 
= (4x^2)*(x^2+2x+1) / (2x^2 + 2x + 1)^2 = [(2x )* (x+1)]^2 / (2x^2 + 2x + 1)^2 
This gives sinA = (2x )* (x+1) / (2x^2 +2x +1)

These would the corresponding values for sinA and cosA respectively.