19: The right inequality:
Denoting the function on the left as f and the one on the right as g one gets

Obviously f ′ < g′ on M = (0,π/2).

The conclusion then follows from

f (0⁺) = 0 = g(0⁺).

The left inequality:
Denoting the function on the left as f and the one on the right as g one gets

Here we do not have f ′ < g′ on M = (0,π/2), so the usual trick did not work out.

Another option is to consider the difference and determine the way it goes using monotonicity.

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