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1 solution
since we need to find maximum value of given expression therefore we actually need to find the minimum value of denominator
so w.r.t. denominator :- (sinA)^2 + (cosA)^2 + 3sinAcosA + 4(cosA)^2
1 + 3/2*sin2A + 2 + 2cos2A
now let 2A = X
so we have,
3 + 3/2*sinX + 2cosX
now for minimum value differentiate and put it eequal to zero
hence we have,
3/2*cosX - 2 sinX = 0
therefore on solving we get TanX = 3/4
now we have the basic 3,4,5 triangle which has 5 as the hypotenuse and 4 and 3 as the base and perpendicular.
therefore since we have sin and cos values in the given expression, therefore they will have minimum values in the third quadrant.
therfore the minimum values are :-
cosX = -4/5 and sinX = -3/5
hence on substituting the values in the given initial expression the maximum value becomes 2