The following equations represent transverse waves
$z_{1} = A \cos (3 K x - ω t)$ …(i)
$z_{2} = A \cos (4 K y - ω t)$ …(ii)
A wave is formed by the superposition of waves (i) and (ii). This wave is

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Traveling in the direction making an angle $\tan^{- 1} \frac{3}{4}$ with +ve x–axis

Traveling in the direction making an angle $π - \tan^{- 1} \frac{3}{4}$ with +ve y–axis

Traveling in the direction making an angle $\tan^{- 1} \frac{4}{3}$ with +ve x–axis

Traveling in the direction making an angle $π + \tan^{- 1} \frac{4}{3}$ with +ve y–axis

answer is B.

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Detailed Solution

Equation of resultant wave is
$Z = z_{1} + z_{2} = 2 A \cos (\frac{3 K x - 4 K y}{2}) \cos (\frac{3 K x + 4 K y}{2} - ω t)$
Equation of wavefront is $\frac{3 K x + 4 K y}{2} - ω t = C$ (C is constant)
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So at any given instant of time the equation of wave front will be
3Kx + 4 Ky = C¢
Wave travels perpendicular to the wave front also with increase in time both x and y should increase, as $\tan θ = \frac{\frac{C'}{3 K}}{\frac{C'}{4 K}} = \frac{4}{3}, θ = \tan^{- 1} \frac{4}{3}$ .