Show that the function f given by f(x) = tan^-1(sinx + cosx), x > 0 is always strictly increasing function in .

f(x) = tan^-1(sinx + cosx)

f(x) is strictly increasing in if f’(x) > 0 when x is between 0 and

let us find f’(x)

differentiating f(x)

Since sin²x + cos² = 1 and 2sinxcosx = sin2x

Consider the numerator cosx – sinx

Observe that from graph cosx > sinx when x is in between 0 and

Blue is cosx and green is sinx

Hence cosx – sinx is positive

Now consider the denominator 2 + sin2x

When x is in sin2x is in that is (0, 1)

Hence 2 + sin2x is also positive

As both numerator and denominator are positive hence f’(x) is also positive that is f’(x) > 0

Hence the function f(x) is strictly increasing in