 Hi and welcome to the session. Let us discuss the following question. Question says, find the general solution for each of the following differential equations. Given differential equation is, sec square x tan y dx plus sec square y tan x dy is equal to 0. Let us now start with the solution. Now given differential equation is, sec square x tan y dx plus sec square y tan x dy is equal to 0. Now subtracting this term from both sides of this differential equation, we get sec square x tan y dx is equal to minus sec square y tan x dy. Now let us name this equation as equation 1. Now separating variables in equation 1, we get sec square x upon tan x dx is equal to minus sec square y upon tan y dy. Now multiplying both the sides of this equation y minus 1, we get sec square y upon tan y dy is equal to minus sec square x upon tan x dx. Now integrating both the sides of this equation, we get integral of sec square y upon tan y with respect to y is equal to negative of integral sec square x dx upon tan x. Now let us name this expression as 2. Now first of all let us find out this integral that is integral of sec square y dy upon tan y. Now we know derivative of tan y is sec square y. So we will solve this integral by substitution method. Now we can write put tan y is equal to t. Now differentiating both the sides with respect to y, we get sec square y is equal to dt upon dy. Now this further implies sec square y dy is equal to dt. Now we can write this given integral is equal to integral of dt upon t. Substituting dt for sec square y dy and t for tan y in this integral, we get this integral is equal to integral of dt upon t. Now using this formula of integration, we can write this integral is equal to log of t plus log c1. Here log c1 represents constant of integration. Now we will substitute tan y for t here and we get log of tan y plus log of c1. So we get integral of sec square y dy upon tan y is equal to log of tan y plus log of c1. Similarly integral of sec square x dx upon tan x is equal to log of tan x plus log of c2. Clearly we can see these two integrals are exactly same except this y has been replaced by x in this integral. Now let us name this expression as 3 and this expression as 4. Now we will substitute values of these two integrals from expression 3 and 4 in expression 2. We know this is the expression 2. Now we can write substituting 3 and 4 into we get log of tan y plus log of c1 is equal to minus log of tan x minus log c2. Now this further implies log of tan y plus log of tan x is equal to minus log c2 minus log c1. Now substituting log c for minus log c2 minus log c1 we get log of tan y plus log of tan x is equal to log c. Now applying this law of logarithms in left hand side of this expression we get log of tan y multiplied by tan x is equal to log c. Now applying this law of logarithms on both sides of this expression we get tan y multiplied by tan x is equal to c. Now the required solution of given differential equation is tan x multiplied by tan y is equal to c. This is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.