Question

Using Lagrange's mean value theorem, prove that (b − a) sec² a < tan b − tan a < (b − a) sec² b
where 0 < a < b < \[\frac{\pi}{2}\] ?

Answer 1

​Consider, the function

\[f\left( x \right) = \tan x, x \in \left[ a, b \right], 0 < a < b < \frac{\pi}{2}\]

Clearly, \[f\left( x \right)\] is continuous on \[\left[ a, b \right]\] and derivable on  \[\left( a, b \right)\] .

Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently,\[c \in \left( a, b \right)\] such that \[f'\left( c \right) = \frac{f\left( b \right) - f\left( a \right)}{b - a}\] .

Now, 

\[f\left( x \right) = \tan x\]\[\Rightarrow\] \[f'\left( x \right) = se c^2 x\],\[f\left( a \right) = \tan a, f\left( b \right) = \tan b\]

\[\therefore\] \[f'\left( c \right) = \frac{f\left( b \right) - f\left( a \right)}{b - a}\]\[\Rightarrow\] \[se c^2 c = \frac{\tan b - \tan a}{b - a} . . . \left( 1 \right)\]

Now, 

\[c \in \left( a, b \right)\]

\[ \Rightarrow a < c < b\]

\[ \Rightarrow se c^2 a < se c^2 c < se c^2 b \left[ \because se c^2 x \text {b is increasing in } \left( 0, \frac{\pi}{2} \right) \right]\]

\[ \Rightarrow se c^2 a < \frac{\tan b - \tan a}{b - a} < se c^2 b \left[ \text { from } \left( 1 \right) \right]\]

\[ \Rightarrow \left( b - a \right)se c^2 a < \tan b - \tan a < \left( b - a \right)se c^2 b\]

Hence proved.