Solution We have
f (x) = tan–1(sin x + cos x), x > 0
Therefore
f ′(x) =
2
1
(cos
sin )
1
(sin
cos )
x
x
x
x
−
+
+
Fig 6 22
Rationalised 2023-24
APPLICATION OF DERIVATIVES
181
= cos
sin
2
xsin 2
x
x
−
+
(on simplification)
Note that 2 + sin 2x > 0 for all x in 0, 4
π Therefore
f ′(x) > 0  if cos x – sin x > 0
or
f ′(x) > 0  if cos x > sin x  or cot x > 1
Now
cot x > 1 if tan x < 1, i e