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Prove 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.

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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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