Question

give the solution of 2-(cosA+sinA)^2=?​

Answer 1

$2 - {(cosa \: + \: sina)}^{2} \\ \\ = 2 - ( {cos}^{2} a + {sin}^{2} a + 2sina \: cosa) \\ \\ = 1 - 2sina \: cosa \\ \\ = {cos}^{2} a + {sin}^{2} a - 2sina \: cosa \\ \\ = {(cosa - sina)}^{2}$

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ \\ {cos}^{2} a + {sin}^{2} a = 1$

hope it helps