Question

If a, b are positive numbers such that a>b, then the minimum value of $a\sec (\theta ) - b\tan (\theta )$ (where, $0 < \theta {\text{;}} < \pi /2$) is
1. $\dfrac{1}{{\sqrt {{a^2} - {b^2}} }}$
2. $\dfrac{1}{{\sqrt {{a^2} + {b^2}} }}$
3. $\sqrt {{a^2} + {b^2}}$
4. $\sqrt {{a^2} - {b^2}}$
5. ${a^2} - {b^2}$

Answer 1

This question involves the use of differentiation. To find the function's minimum or maximum value, we need to find the critical points using differentiation and double differentiate the final equation to confirm whether the point will give maxima or minima.
For minima the conditions are: $\Rightarrow \dfrac{{dy}}{{d\theta }} = 0$,$\Rightarrow \dfrac{{{d^2}y}}{{d{\theta ^2}}} > 0$
Also, for solving this question one needs to be aware about manipulating trigonometric functions in terms of the ratio of the sides of the triangle, i.e.,
$\sin (\theta ) = \dfrac{P}{H}$
$\cos (\theta ) = \dfrac{B}{H}$
$\tan (\theta ) = \dfrac{P}{B}$
$\sec (\theta ) = \dfrac{H}{B}$
$\cos ec(\theta ) = \dfrac{H}{P}$
$\cot (\theta ) = \dfrac{B}{P}$, where P represents the perpendicular of the triangle, B represents Base of the triangle and H represents hypotenuse of the triangle.

Complete answer:
We have, $y = a\sec (\theta ) - b\tan (\theta )$, lets evaluate,$\dfrac{{dy}}{{d\theta }}$,$\Rightarrow \dfrac{{dy}}{{d\theta }} = a\sec (\theta )\tan (\theta ) - b{\sec ^2}(\theta )$
$\Rightarrow \dfrac{{dy}}{{d\theta }} = \sec (\theta )(a\tan (\theta ) - b\sec (\theta )) - - - - - \left( 1 \right)$
To find the critical points,
$\Rightarrow \dfrac{{dy}}{{d\theta }} = 0$
$\Rightarrow (a\tan (\theta ) - b\sec (\theta )) = 0$ , as $\sec (\theta ) \ne 0$ in $0 < \theta {\text{;}} < \pi /2$
$\Rightarrow \sin (\theta ) = \dfrac{b}{a}$
Now, to check whether this critical point returns the minima or the maxima we need to again differentiate the above equation and put $\sin (\theta ) = \dfrac{b}{a}$.
So, $\dfrac{{{d^2}y}}{{d{\theta ^2}}} = \dfrac{d}{{d\theta }}\left[ {\sec (\theta )(a\tan (\theta ) - b\sec (\theta ))} \right]$
Using product rule of differentiation, $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right)$, we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{\theta ^2}}} = \sec (\theta )\dfrac{d}{{d\theta }}\left[ {(a\tan (\theta ) - b\sec (\theta ))} \right] + (a\tan (\theta ) - b\sec (\theta ))\dfrac{d}{{d\theta }}\left[ {\sec (\theta )} \right]$
Now, substituting the derivatives of basic trigonometric functions, we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{\theta ^2}}} = \sec \theta \left[ {a{{\sec }^2}\theta - b\sec \theta \tan \theta } \right] + (a\tan \theta - b\sec \theta )\sec \theta \tan \theta$
Now, substituting the value of $\theta$ as ${\sin ^{ - 1}}\left( {\dfrac{b}{a}} \right)$.
Since, $\sin (\theta ) = \dfrac{P}{H} = \dfrac{b}{a}$, where P refers to perpendicular and H represents to hypotenuse.
Let P be b and H be a.
Then, using Pythagoras theorem, we get,
${P^2} + {B^2} = {H^2}$
$\Rightarrow {b^2} + {B^2} = {a^2}$
So, $B = \sqrt {{a^2} - {b^2}}$
Therefore, $\tan (\theta ) = \dfrac{P}{B} = \dfrac{b}{{\sqrt {{a^2} - {b^2}} }}$ and $\sec (\theta ) = \dfrac{a}{{\sqrt {{a^2} - {b^2}} }}$.
So, we get,
\Rightarrow \dfrac{{{d^2}y}}{{d{\theta ^2}}}\left( {\theta = {{\sin }^{ - 1}}\dfrac{b}{a}} \right) = \dfrac{a}{{\sqrt {{a^2} - {b^2}} }}\left[ {a{{\left( {\dfrac{a}{{\sqrt {{a^2} - {b^2}} }}} \right)}^2} - b\dfrac{a}{{\sqrt {{a^2} - {b^2}} }}\dfrac{b}{{\sqrt {{a^2} - {b^2}} }}} \right] + \left( {a\dfrac{b}{{\sqrt {{a^2} - {b^2}} }} - b\dfrac{a}{{\sqrt {{a^2} - {b^2}} }}} \right)\dfrac{a}{{\sqrt {{a^2} - {b^2}} }}\dfrac{b}{{\sqrt {{a^2} - {b^2}} }}$$$$ \Rightarrow \dfrac{{{d^2}y}}{{d{\theta ^2}}} = \dfrac{a}{{\sqrt {{a^2} - {b^2}} }}\left[ {a\left( {\dfrac{{{a^2}}}{{{a^2} - {b^2}}}} \right) - b\dfrac{{ab}}{{{a^2} - {b^2}}}} \right] + \left( {\dfrac{{ab}}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{ab}}{{\sqrt {{a^2} - {b^2}} }}} \right)\left( {\dfrac{{ab}}{{{a^2} - {b^2}}}} \right)
Simplifying the expression further, we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{\theta ^2}}} = \dfrac{{{a^2}}}{{\sqrt {{a^2} - {b^2}} }}\left[ {\dfrac{{{a^2}}}{{{a^2} - {b^2}}} - \dfrac{{{b^2}}}{{{a^2} - {b^2}}}} \right] + 0$
After all the calculations we get, $\Rightarrow \dfrac{{{d^2}y}}{{d{\theta ^2}}} > 0,$
Therefore, the function returns a minima at $\sin (\theta ) = \dfrac{b}{a}$
Now, to find the minimum value replace $\theta$ by ${\sin ^{ - 1}}(\dfrac{b}{a})$
Also, we have, $\sec (\theta ) = \dfrac{a}{{\sqrt {{a^2} - {b^2}} }}$ and $\tan (\theta ) = \dfrac{b}{{\sqrt {{a^2} - {b^2}} }}$.
Now, replacing the values of functions from the above equations in the given equation $y = f(x)$
, i.e., in $y = a\sec (\theta ) - b\tan (\theta )$, we get,
After replacing the values we get,
$\Rightarrow y = \dfrac{{a \times a}}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{b \times b}}{{\sqrt {{a^2} - {b^2}} }}$
Solving the numerators together as the denominator is same,
$\Rightarrow y = \dfrac{{{a^2} - {b^2}}}{{\sqrt {{a^2} - {b^2}} }}$
And finally dividing the numerator by denominator, which is nothing but the square of the denominator,
$\Rightarrow y = \sqrt {{a^2} - {b^2}}$
Thus, option (4) is the correct answer.

Note:
We use the maxima and minima concept by two methods: the first derivative test and the second derivative test. The first derivative test helps in finding the local extremum points of the function. The second derivative test involves finding the local extremum points and then finding out which of them is local minima and which is local maxima. If the second derivative of a function at a local extremum point is positive, it means that the point corresponds to minima and if it is negative, it corresponds to maxima.