Solution of the differential equation tan y sec² x dx + tan x sec² y dy + 0 is .

The correct option is tan x.tan y = k.

Step-by-Step Explanation:
We are given the first-order differential equation:
$$tany{sec}^{2}xdx+tanx{sec}^{2}ydy=0$$

This equation is a variable separable differential equation. To separate the variables, we divide the entire equation by the product $tanxtany$:
$$\frac{{sec}^{2}x}{tanx}dx+\frac{{sec}^{2}y}{tany}dy=0$$

Now that the variables are separated, we integrate both sides of the equation:
$$\int \frac{{sec}^{2}x}{tanx}dx+\int \frac{{sec}^{2}y}{tany}dy=C$$

To evaluate these integrals, we can use the method of substitution. For the first term, let:
$$u=tanx$$
Differentiating both sides with respect to $x$ gives:
$$du={sec}^{2}xdx$$
Substituting these into the first integral:
$$\int \frac{1}{u}du=ln\left|u\right|=ln|tanx|$$

Applying the exact same substitution logic for the second integral with respect to $y$, we get:
$$\int \frac{{sec}^{2}y}{tany}dy=ln|tany|$$

Substituting these results back into our integrated equation, and expressing the constant of integration as $lnk$ for convenience, we have:
$$ln|tanx|+ln|tany|=lnk$$

Using the logarithmic property $lnA+lnB=ln(A\cdot B)$, we can combine the terms on the left-hand side:
$$ln|tanx\cdot tany|=lnk$$

By taking the exponential of both sides, we eliminate the logarithms:
$$tanx\cdot tany=k$$

Thus, the general solution of the given differential equation is indeed $tanx\cdot tany=k$.