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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

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### 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?

#### APPEARS IN

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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