Question

4sinA + 3cosA =5 then find (3 sinA - 4cosA) whole square

Answer 1

$4si n + 3cos = 5 \\ then \\ 3sin - 4cos \\ = \sqrt{ {4}^{2} + {3}^{2} - {5}^{2} } \\ = 0 \\ hence \: (3sin - 4cos) = 0$
and
squaring both sides
${(3sin - 4cos)}^{2} = 0$