Let's solve the equation:
[tex]\[
\sec 40^{\circ} + \sqrt{3} \operatorname{cosec} 40^{\circ} = 4
\][/tex]
We start by recalling the definitions of the secant and cosecant functions:
[tex]\[
\sec x = \frac{1}{\cos x} \quad \text{and} \quad \csc x = \frac{1}{\sin x}
\][/tex]
Substituting these definitions into the given equation, we get:
[tex]\[
\frac{1}{\cos 40^{\circ}} + \sqrt{3} \cdot \frac{1}{\sin 40^{\circ}} = 4
\][/tex]
Combining the terms over a common denominator, which is [tex]\(\cos 40^{\circ} \sin 40^{\circ}\)[/tex]:
[tex]\[
\frac{\sin 40^{\circ} + \sqrt{3} \cos 40^{\circ}}{\cos 40^{\circ} \sin 40^{\circ}} = 4
\][/tex]
Multiplying both sides by [tex]\(\cos 40^{\circ} \sin 40^{\circ}\)[/tex] to clear the fraction:
[tex]\[
\sin 40^{\circ} + \sqrt{3} \cos 40^{\circ} = 4 \cos 40^{\circ} \sin 40^{\circ}
\][/tex]
We can use the double-angle identity for sine on the right side:
[tex]\[
\sin 40^{\circ} + \sqrt{3} \cos 40^{\circ} = 2 \sin 40^{\circ} \cos 40^{\circ}
\][/tex]
This reduces to:
[tex]\[
\sin 40^{\circ} + \sqrt{3} \cos 40^{\circ} = 2 \sin 80^{\circ}
\][/tex]
Since [tex]\(\sin 80^\circ = \cos 10^\circ\)[/tex], we have:
[tex]\[
\sin 40^{\circ} + \sqrt{3} \cos 40^{\circ} = \sin 40^{\circ}
\][/tex]
Subtracting [tex]\(\sin 40^\circ\)[/tex] from both sides of the equation:
[tex]\[
\sqrt{3} \cos 40^{\circ} = 0
\][/tex]
This implies:
[tex]\[
\cos 40^{\circ} = 0
\][/tex]
However, we know that [tex]\(\cos 40^{\circ} \neq 0\)[/tex] because [tex]\(40^\circ\)[/tex] is not an angle for which cosine is zero. Therefore, there is no solution to this equation.
The numerical result confirms that there are no solutions:
[tex]\[
\boxed{\text{No solution}}
\][/tex]
