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**Hint:**We need to simplify such equation using trigonometric functions

Sum of the trigonometric functions can be formulated as:

\[Cos(A + B) + \operatorname{Cos} (A - B) = 2\operatorname{Cos} A\operatorname{Cos} B\]

We know that the value of a \[\operatorname{Sin} \theta + B\operatorname{Cos} \theta \] lies between: \[ - \sqrt {{a^2} + {b^2}} \leqslant a\operatorname{Sin} \theta + b\cos \theta \leqslant \sqrt {{a^2} + {b^2}} \].

**Complete step-by- step solution:**Given that

\[\sin x + \cos x = 2 \ldots (1)\]

Taking square both sides

$\Rightarrow$\[{(\sin x + \cos x)^2} = {a^2}\]

As we know \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]

$\Rightarrow$\[{\sin ^2}x + {\cos ^2}x + 2\sin x\,\cos x = {a^2}\]

But, \[{\sin ^2}x + {\cos ^2}x = 1\]

$\Rightarrow$\[1 + 2\,\sin x.\,\cos x = {a^2}\]

$\Rightarrow$\[\sin x\,\cos x = \dfrac{{{a^2} - 1}}{2} \ldots (2)\]

Using equation (1)

And applying \[{(a - b)^2} = {(a + b)^2} = 4ab\]

$\Rightarrow$\[{(\sin x - \cos x)^2} = {(\sin x + \cos x)^2} - 4\sin x\,\cos x\]

But the values from equation (1) & (2).

\[{(\sin x - \cos x)^2} = {a^2} - 4\left( {\dfrac{{{a^2} - 1}}{2}} \right)\]

\[ = {a^2} - 2({a^2} - 1)\]

\[ = {a^2} - 2{a^2} + 2\]

\[ = - {a^2} + 2\]

\[{(\sin x - \cos x)^2} = 2 - {a^2}\]

Taking square root both sides

$\Rightarrow$\[\sin x - \,\cos x = \pm \sqrt {2 - {a^2}} \]

$\Rightarrow$\[\,\left| {\sin x - \cos x} \right| = \sqrt {2 - {a^2}} \]

Hence the value of \[\left| {\sin x - \,\cos x} \right|\] would be positive square root of \[(2 - {a^2})\] i.e. \[\sqrt {2 - {a^2}} \]

**Note:**Modulus of trigonometric functions and periods, we know that modulus operation on function converts negative function values to positive function values with equal magnitude. As such, we draw a graph of the modulus function by taking a mirror image of the corresponding core graph in x-axis.

Recall that in its basic form \[\,f(x) = |x|,\,\]the absolute value function is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems.

this kind of question using \[{\sin ^2}x + {\cos ^2}x = 1\], \[{(a - b)^2} = {(a + b)^2} - 4ab\], and \[{(a + b)^2} = {a^2} + {b^2} + 2ab\].

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