We have to find the maximum value of 1−4sin2xcos2x Let f(x)=1−4sin2xcos2x =1−(2sinxcosx)2 =1−sin22x(∵sin2θ=2sinθcosθ) For maximum value of f(x), sin 2x should be minimum As we know that, Minimum value of sin2θ is 0 Therefore, Maximum value of f(x) = 1 – 0 = 1