Prove that (b-a)/(cos²a) < tan b- tan a < (b-a)/(cos²b)

Prove that $\dfrac{b-a}{\cos^{2}a} \lt \tan b-\tan a \lt \dfrac{b-a}{\cos^{2}b}$ for $0\lt a\lt b\lt \dfrac{\pi}{2}$.

Step 1: Apply Lagrange's theorem

On $[a,b]\subset\left(0,\tfrac{\pi}{2}\right)$ the function $\tan x$ is continuous (since $\cos x\neq0$ there) and differentiable, with $\bigl(\tan x\bigr)' = \dfrac{1}{\cos^{2}x}$. Hence for some $\xi\in(a,b)$

\[ \tan b-\tan a = \frac{b-a}{\cos^{2}\xi}. \tag{1} \]

Step 2: The derivative is increasing

Put $h(x) = \dfrac{1}{\cos^{2}x}$. Then

\[ h'(x) = \frac{2\sin x}{\cos^{3}x} > 0 \ \text{ on } \left(0,\frac{\pi}{2}\right), \]

because both $\sin x$ and $\cos x$ are positive there. So $h$ is strictly increasing.

Step 3: Bracket the unknown point

From $a\lt \xi\lt b$ and the monotonicity of $h$: $\frac{1}{\cos^{2}a} \lt \frac{1}{\cos^{2}\xi} \lt \frac{1}{\cos^{2}b}$.

Multiplying by $b-a>0$ and using (1) gives the required chain.

Step 4: Verification with $a = \dfrac{\pi}{6},\ b = \dfrac{\pi}{4}$

\[b-a = \frac{\pi}{12} \approx 0.261799; \qquad \text{left } = \frac{\pi/12}{3/4} = \frac{\pi}{9} \approx 0.34907\]

\[\text{middle } = 1-\frac{1}{\sqrt3} \approx 0.42265; \qquad \text{right } = \frac{\pi/12}{1/2} = \frac{\pi}{6} \approx 0.52360\]

\[0.34907 \lt 0.42265 \lt 0.52360 \;\;\checkmark\]