Question

prove that sinA - cosA the whole square + 2 sinA cosA is equal to 1​

Answer 1

Step-by-step explanation:

(sinA-cosA)²+2sinAcosA=1

L.H.S= (sinA-cosA)²+2sinAcosA

(sin²A +cos²A -2sinAcosA) +2sinAcosA

sin²A + cos²A-2sinAcosA+2sinAcosA

1+0

1

L.H.S=R.H.S

Answer 2

Step-by-step explanation:

(sin A -cos A)^2 + 2 sinA cosA = 1

sin A -cos A)^2 + 2 sinA cosA = 1sin^2 A + cos^2A -2 sinA cosA + 2 sinA cosA = 1

sin A -cos A)^2 + 2 sinA cosA = 1sin^2 A + cos^2A -2 sinA cosA + 2 sinA cosA = 1sin^2 A + cos^2A = 1

sin A -cos A)^2 + 2 sinA cosA = 1sin^2 A + cos^2A -2 sinA cosA + 2 sinA cosA = 1sin^2 A + cos^2A = 11 = 1 (sin^2 A + cos^2A = 1)

Hope it is helpful for you ✌❤