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Solution for Using Lagrange'S Mean Value Theorem, Prove that (B − A) Sec2 a < Tan B − Tan a < (B − A) Sec2 B Where 0 < a < B < π 2 ? - CBSE (Commerce) Class 12 - Mathematics

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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 < \[\frac{\pi}{2}\] ?

Solution

​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.

  Is there an error in this question or solution?
Solution for 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 < π 2 ? concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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