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(b - a)/(1 - b2) < tan–1 b – tan–1 a < (b - a)/(1 + a2), if 0 < a < b and deduce that  π/4 + 3/25 < tan-1 4/3 < π/4 + 1/6.

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Let f (x) = tan–1x is continuous in [a, b], a > 0.

Hence f ′ (x) = 1/(1 + x2),  f(x) is differentiable in (a, b).

Applying Lagrange’s theorem, we get

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