The correct option is D g(x) has local maximum at x=3π4
g(x)=f(tanx)+f(cotx) ∀x∈(π2,π)
g′(x)=f′(tanx)sec2x−f′(cotx)(cosec2x)
For increasing, g′(x)>0
f′′(x)<0⇒f′(x) is decreasing
∵tanx<cotx∀x∈(π2,3π4)∴f′(tanx)>f′(cotx)
Also, sec2x>cosec2 ∀x∈(π2,3π4)
∴g′(x)>0⇒g(x) is increasing in
(π2,3π4)
Similarly, g(x) is decreasing in (3π4,π)
Also, g′(3π4)=0
At x=3π4, sign of g′(x) changes from positive to negative
So, g(x) has local maximum at x=3π4