Question

The minimum value of $\left| {\sin x + \cos x + \tan x + \sec x + \cos ec x + \cot x} \right|$ is
A) $2\sqrt 2 - 1$
B) $2\sqrt 2 + 1$
C) $\sqrt 2 - 1$
D) $\sqrt 2 + 1$

Answer 1

Here, we will first convert all these functions in terms of the sine and cosine. Then we will add them together and simplify it further to find the minimum value of the sum. Trigonometric ratios are defined as the ratios of any two sides of a right angled triangle.

Formula Used:
We will use the following Formula:

1. Trigonometric ratio:$\tan x = \dfrac{{\sin x}}{{\cos x}};\sec x = \dfrac{1}{{\cos x}};\cos ecx = \dfrac{1}{{\sin x}};\cot x = \dfrac{{\cos x}}{{\sin x}}$
2. Trigonometric identities: ${\sin ^2}x + {\cos ^2}x = 1$
3. The square of the sum of two numbers is given by the formula: ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
4. The difference of square of two numbers is given by the algebraic identity $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$
5. Trigonometric formula: $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
6. \left| x \right| = \left\\{ \begin{array}{l}x,{\rm{ }}if{\rm{ }}x \ge 0\\\ - x,{\rm{ }}if{\rm{ }}x < 0\end{array} \right.

Complete step by step solution:
We are given with a trigonometric function $\left| {\sin x + \cos x + \tan x + \sec x + \cos ecx + \cot x} \right|$
Let $\left| {\sin x + \cos x + \tan x + \sec x + \cos ecx + \cot x} \right| = y$
Now, we have to represent all the terms in terms of sine and cosine.
Now, substituting $\tan x = \dfrac{{\sin x}}{{\cos x}},\sec x = \dfrac{1}{{\cos x}},\cos ecx = \dfrac{1}{{\sin x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$ in the given expression, we get
$\Rightarrow y = \sin x + \cos x + \dfrac{{\sin x}}{{\cos x}} + \dfrac{1}{{\cos x}} + \dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}}$

By grouping the terms and taking L.C.M., we get
$\Rightarrow y = \sin x + \cos x + \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}} + \dfrac{1}{{\sin x}} + \dfrac{1}{{\cos x}}$
$\Rightarrow y = \sin x + \cos x + \dfrac{{\sin x}}{{\cos x}} \times \dfrac{{\sin x}}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}} \times \dfrac{{\cos x}}{{\cos x}} + \dfrac{1}{{\sin x}} \times \dfrac{{\cos x}}{{\cos x}} + \dfrac{1}{{\cos x}} \times \dfrac{{\sin x}}{{\sin x}}$
Adding the like terms, we get
$\Rightarrow y = \sin x + \cos x + \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\cos x\sin x}} + \dfrac{{\cos x + \sin x}}{{\cos x\sin x}}$
Now, by using the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$, we get
$\Rightarrow y = \sin x + \cos x + \dfrac{{1 + \cos x + \sin x}}{{\cos x\sin x}}$ ………………………..$\left( 1 \right)$
Using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, we can write
${\left( {\sin x + \cos x} \right)^2} = {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x$
$\Rightarrow {\left( {\sin x + \cos x} \right)^2} = 1 + 2\sin x\cos x$
Rewriting the equation, we get
$\Rightarrow \sin x\cos x = \dfrac{{{{\left( {\sin x + \cos x} \right)}^2} - 1}}{2}$
Substituting $\sin x\cos x = \dfrac{{{{\left( {\sin x + \cos x} \right)}^2} - 1}}{2}$ in equation $\left( 1 \right)$, we get
$\Rightarrow y = \sin x + \cos x + \dfrac{{1 + \cos x + \sin x}}{{\dfrac{{{{\left( {\sin x + \cos x} \right)}^2} - 1}}{2}}}$
Substituting $\sin x + \cos x = t$ in the above equation, we get
$\Rightarrow \sin x + \cos x + \dfrac{{1 + \cos x + \sin x}}{{\dfrac{{{{\left( {\sin x + \cos x} \right)}^2} - 1}}{2}}} = t + \dfrac{{1 + t}}{{\dfrac{{{t^2} - 1}}{2}}}$
$\Rightarrow \sin x + \cos x + \dfrac{{1 + \cos x + \sin x}}{{\dfrac{{{{\left( {\sin x + \cos x} \right)}^2} - 1}}{2}}} = t + \dfrac{{2\left( {1 + t} \right)}}{{{t^2} - 1}}$
The difference of square of two numbers is given by the algebraic identity $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$
$\Rightarrow \sin x + \cos x + \dfrac{{1 + \cos x + \sin x}}{{\dfrac{{{{\left( {\sin x + \cos x} \right)}^2} - 1}}{2}}} = t + \dfrac{{2\left( {1 + t} \right)}}{{\left( {1 + t} \right)\left( {t - 1} \right)}}$
$\Rightarrow \sin x + \cos x + \dfrac{{1 + \cos x + \sin x}}{{\dfrac{{{{\left( {\sin x + \cos x} \right)}^2} - 1}}{2}}} = t + \dfrac{2}{{\left( {t - 1} \right)}}$………………………..$\left( 2 \right)$
We know that $t = \sin x + \cos x$.
Multiplying and dividing by $\sqrt 2$on both the sides, we get
$\Rightarrow t = \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\sin x + \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\cos x = \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x} \right)$
$\Rightarrow t = \sqrt 2 \left( {\cos \dfrac{\pi }{4}\sin x + \sin \dfrac{\pi }{4}\cos x} \right)$
Now, by using trigonometric formula $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$, we get
$\Rightarrow t = \sqrt 2 \left( {\sin \left( {x + \dfrac{\pi }{4}} \right)} \right)$
Thus , by this identity, we will get $- \sqrt 2 \le t \le \sqrt 2$ .
Substituting $y = \left| {\sin x + \cos x + \tan x + \sec x + \cos ecx + \cot x} \right| = t + \dfrac{2}{{\left( {t - 1} \right)}}$
$\Rightarrow y = \left| {t + \dfrac{2}{{t - 1}}} \right| = \left| {t - 1 + \dfrac{2}{{t - 1}} + 1} \right|$
Now, by using the property \left| x \right| = \left\\{ \begin{array}{l}x,{\rm{ }}if{\rm{ }}x \ge 0\\\ - x,{\rm{ }}if{\rm{ }}x < 0\end{array} \right. , we get
Now, considering $1 < a < \sqrt 2 \Rightarrow 0 < a - 1$in this case the value is positive.
$\Rightarrow y = t - 1 + \dfrac{2}{{t - 1}} + 1$
We know that Arithmetic mean$\ge$ Geometric mean. Thus, we will get
$\Rightarrow \dfrac{{t - 1 + \dfrac{2}{{t - 1}}}}{2} \ge \sqrt {\left( {t - 1} \right) \times \dfrac{2}{{t - 1}}}$
$\Rightarrow \dfrac{{t - 1 + \dfrac{2}{{t - 1}}}}{2} \ge \sqrt 2 \Rightarrow t - 1 + \dfrac{2}{{t - 1}} \ge 2\sqrt 2$
$\Rightarrow t - 1 + \dfrac{2}{{t - 1}} + 1 \ge 2\sqrt 2 + 1$
Thus, we get
$\Rightarrow y \ge 2\sqrt 2 + 1$ which is positive…………………………………………………..$\left( 3 \right)$
Now, considering $- 1 < a < \sqrt 2 \Rightarrow a - 1 < 0$ in this case the value is negative
$\Rightarrow y = t - 1 + \dfrac{2}{{t - 1}} + 1$
$\Rightarrow y = \left| { - 1\left( {1 - t + \dfrac{2}{{1 - t}} - 1} \right)} \right|$
$\Rightarrow y = \left| { - 1} \right|\left| {\left( {1 - t + \dfrac{2}{{1 - t}} - 1} \right)} \right|$
$\Rightarrow y = \left| {\left( {1 - t + \dfrac{2}{{1 - t}} - 1} \right)} \right|$
We know that Arithmetic mean $\ge$ Geometric mean. Thus, we will get
$\Rightarrow \dfrac{{1 - t + \dfrac{2}{{1 - t}}}}{2} \ge \sqrt {\left( {1 - t} \right) \times \dfrac{2}{{1 - t}}}$
$\Rightarrow \dfrac{{1 - t + \dfrac{2}{{1 - t}}}}{2} \ge \sqrt 2 \Rightarrow 1 - t + \dfrac{2}{{1 - t}} \ge 2\sqrt 2$
$\Rightarrow 1 - t + \dfrac{2}{{1 - t}} - 1 \ge 2\sqrt 2 - 1$
Thus, we get
$\Rightarrow y \ge 2\sqrt 2 - 1$ which is positive………………………………………………$\left( 4 \right)$
Since the value has to be minimum, from $\left( 3 \right)$ and $\left( 4 \right)$, we get $2\sqrt 2 - 1 < 2\sqrt 2 + 1$
Thus the minimum value is $2\sqrt 2 - 1$

Therefore, the minimum value of $\left| {\sin x + \cos x + \tan x + \sec x + \cos ecx + \cot x} \right|$ is $2\sqrt 2 - 1$.

Note:
We can find the limits by using the trigonometric identity and values. So, it becomes important to remember all the basic identities and values. We have found the minimum value using the relation between arithmetic mean and geometric mean. Arithmetic mean or average is defined as the sum of numbers divided by the quantity of numbers, Geometric mean of two numbers is defined as the square root of their product. We should also remember that the Arithmetic mean should never be less than the geometric mean.