Given, the function, f(x)=tan−1(sinx+cosx) in [0,2π]
f′(x)=1+(sinx+cosx)2cosx−sinx
=1+sin2xcosx−sinx
f′(x)=0
cosx−sinx=0
tanx=1
x=4π
Then, 1≤sinx+cosx≤2x∈[4π,tan−12]
M=tan−1(2)
m=tan−1(1)
M−m=tan−1(2)−tan−1(1)
M−m=tan−1(1+22−1)
M−m=tan−1(3−22)
tan(M−m)=3−22