Answer :

To solve the equation [tex]\(\frac{1}{\cos x} + \frac{1}{\sin x} = \sqrt{2} \sin \left(x + \frac{\pi}{4}\right)\)[/tex], we can simplify it using trigonometric identities and algebraic manipulations. Let's proceed step-by-step:

1. Rewrite the Right-Hand Side Using a Trigonometric Identity:

We know from trigonometric identities that:
[tex]\[ \sin \left(x + \frac{\pi}{4}\right) = \sin x \cos \left(\frac{\pi}{4}\right) + \cos x \sin \left(\frac{\pi}{4}\right) \][/tex]

Since [tex]\(\cos \left(\frac{\pi}{4}\right) = \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)[/tex], we substitute:

[tex]\[ \sin \left(x + \frac{\pi}{4}\right) = \sin x \cdot \frac{\sqrt{2}}{2} + \cos x \cdot \frac{\sqrt{2}}{2} \][/tex]

Simplifying this, we get:

[tex]\[ \sin \left(x + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} (\sin x + \cos x) \][/tex]

2. Substitute back into the Original Equation:

The original equation is:
[tex]\[ \frac{1}{\cos x} + \frac{1}{\sin x} = \sqrt{2} \sin \left(x + \frac{\pi}{4}\right) \][/tex]

Substituting the identity we derived:

[tex]\[ \frac{1}{\cos x} + \frac{1}{\sin x} = \sqrt{2} \cdot \frac{\sqrt{2}}{2} (\sin x + \cos x) \][/tex]

Simplify the right-hand side:

[tex]\[ \frac{1}{\cos x} + \frac{1}{\sin x} = (\sin x + \cos x) \][/tex]

3. Simplify the Equation:

We recognize that:
[tex]\[ \frac{1}{\cos x} = \sec x \quad \text{and} \quad \frac{1}{\sin x} = \csc x \][/tex]

Therefore:

[tex]\[ \sec x + \csc x = \sin x + \cos x \][/tex]

4. Equality Verification:

Notice that for the above equation to hold true, [tex]\(\sec x + \csc x\)[/tex] and [tex]\(\sin x + \cos x\)[/tex] must be defined. This would be true where [tex]\(\sin x \neq 0\)[/tex] and [tex]\(\cos x \neq 0\)[/tex].

Since both sides of the simplified equation match, the equation:

[tex]\[ \frac{1}{\cos x} + \frac{1}{\sin x} = \sqrt{2} \sin \left(x + \frac{\pi}{4}\right) \][/tex]

holds for all [tex]\(x\)[/tex] where [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex] are defined and non-zero. Therefore,

[tex]\[ \boxed{\text{The equation holds true for all } x \text{ where } \sin x \text{ and } \cos x \text{ are defined.}} \][/tex]