Show that the function f(x) = cot–1 (sin x + cos x) is decreasing on (0, π/4) and increasing on (π/4, π/2).

Given:- Function f(x) = cot–1 (sin x + cos x)


Theorem:- Let f be a differentiable real function defined on an open interval (a,b).


(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)


(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)


Algorithm:-


(i) Obtain the function and put it equal to f(x)


(ii) Find f’(x)


(iii) Put f’(x) > 0 and solve this inequation.


For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.


Here we have,


f(x) = cot–1 (sin x + cos x)






Now, as given



Cosx – sinx< 0 ; as here cosine values are smaller than sine values for same angle



f’(x) < 0


hence, Condition for f(x) to be decreasing


Thus f(x) is decreasing on interval


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