Question

If sinA+cosA=1 then prove that sinA-cosA=±1

Answer 1

Answer:

sin2A=0

Step-by-step explanation:

sinA+cosA=1

∴(sinA+cosA)2=12

∴sin2A+cos2A+2sinAcosA=1

∴1+sin2A=1

∴sin2A=0

Answer 2

Answer:

Step-by-step explanation:

Given that:

$sin A + cos A =1\\\\Squaring \ both \ sides \ we \ get\\\\\Rightarrow sin^2A + Cos^2A +2sinAcosA =1---------(1)\\\\\Rightarrow 1+2sinAcosA=1\\\\\Rightarrow 2sinAcosA =0\\\\Now \ from\ (1)\\\\sin^A + cos^2A +2sinAcosA=1\\\\Subtracting \ 4sinAcosA from\ both \ sides \\\\sin^A + cos^2A +2sinAcosA-4sinAcosA = 1 - 4sinAcosA\\\\\Rightarrow sin^A + cos^2A -2sinAcosA =1-0\\\\\Rightarrow (sinA -cos A)^2 =1\\\\\Rightarrow sin A-cos A = \pm 1\\\\Hence \ proved$