Answer

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**Hint:**Here it is a trigonometric question. We will solve this by converting sec x into tan form or vice versa. And if we are given a function f(x) then the minimum value will be obtained by putting f’(x)=0 and for more than one value, f’’(x)>0 will give minimum and f’’(x)<0 will give maximum value.

**Complete step-by-step answer:**

Step-1

Let f(x) =$a\sec x - b\tan x$

Differentiating the above function we get,

$f'(x) = \dfrac{d}{{dx}}(a\sec x - b\tan x)$

Step-2

Solving this we get,

$f'(x) = \dfrac{d}{{dx}}a\sec x - \dfrac{d}{{dx}}b\tan x$

Or, $f'(x) = a\dfrac{d}{{dx}}\sec x - b\dfrac{d}{{dx}}\tan x$

Or, $f'(x) = a\sec x \cdot \tan x - b{\sec ^2}x$

Step-3

For the minimum value,

f’(x)=0

Step-4

Hence, $a\sec x \cdot \tan x - b{\sec ^2}x = 0$

$ \Rightarrow \sec x(a\tan x - b\sec x) = 0$

Step-6

sec x can never be equal to 0

Then, a tan x – b sec x = 0

Step-7

Cancelling cos x on both the side we get

a sin x = b

Or, sin x =b/a

Step-8

We know that in a triangle, sin x = perpendicular/hypotenuse

Here p = b and h = a

Therefore, base $b = \sqrt {{h^2} - {p^2}} $

$ \Rightarrow b = \sqrt {{a^2} - {b^2}} $

$\therefore \sec x = \dfrac{a}{{\sqrt {{a^2} - {b^2}} }}and\tan x = \dfrac{b}{{\sqrt {{a^2} - {b^2}} }}$

Step-9

Putting these values in f(x) =$a\sec x - b\tan x$we get,

$ \Rightarrow f'(x) = \sqrt {{a^2} - {b^2}} $

$ \Rightarrow f'(x) = \dfrac{{{a^2} - {b^2}}}{{\sqrt {{a^2} - {b^2}} }}$

$ \Rightarrow f'(x) = \sqrt {{a^2} - {b^2}} $

Step-10

Hence it is proved that the minimum value of $a\sec x - b\tan x$is $\sqrt {{a^2} - {b^2}} $where a and b is positive and a>b.

**Note:**Sec value can never be zero.

To get the minimum value f’(x) must be equal to 0.

You can also solve the same problem by converting sec x into tan form and then by differentiating it with respect to x.

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