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If tanA=2(x+1)/2x+1 find sinA and cosA in terms of x
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 = 1/secA = (2x + 1) / (2x^2 + 2x + 1) --------- [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) --------- [2]
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