Question

Anyone help me solve this​

Answer 1

Answer:

$(1-sinA-cosA)^2 = 2(1-sinA)(1-cosA)\\Solution,\\L.H.S. = (1-sinA-cosA)^2 = [(1-sinA)-cosA]^2\\~~~~~~~~~~=(1-sinA)^2-2cosA(1-sinA)+cos^2A\\~~~~~~~~~~=1-2sinA+sin^2A-2cosA(1-sinA)+cos^2A\\~~~~~~~~~~=1-2sinA+sin^2A+cos^2A-2cosA(1-sinA)\\~~~~~~~~~~=1-2sinA+1-2cosA(1-sinA)\\~~~~~~~~~~=2-2sinA-2cosA(1-sinA)\\~~~~~~~~~~=2(1-sinA)-2cosA(1-sinA)\\~~~~~~~~~~=(1-sinA)(2-2cosA)= 2(1-sinA)(1-cosA)$