 Hi and welcome to the session. Let us discuss the following question. Question says discuss the continuity of cosine, cosecant, secant and cotangent functions. First of all let us understand that function f is continuous at x equal to a if function is defined at x equal to a or we can say f a exist. And second condition is limit of the function is equal to value of the function at x equal to a. This is the key idea to solve the given question. Let us now start the solution. First of all let us discuss the continuity of cosine function. First of all let us consider any real number a. Now let fx is equal to cos x then limit of x tending to a fx is equal to limit of x tending to a cos x. Now put x equal to a plus h then as x tends to a h tends to 0. Now this limit is equal to limit of h tending to 0 cos a plus h. Now this is equal to limit of h tending to 0 cos a cos h minus sin a sin h. We know cos a plus b is equal to cos a cos b minus sin a sin b. So here we have applied the formula of cos a plus b. Now this is further equal to cos a limit of h tending to 0 cos h minus sin a limit of h tending to 0 sin h. Now this is further equal to cos a multiplied by cos 0 minus sin a multiplied by sin 0. Now we know cos 0 is equal to 1 and sin 0 is equal to 0. So we can write it equal to cos a multiplied by 1 minus sin a multiplied by 0. This is equal to cos a. So we get limit of x tending to a fx equal to cos a. Now let us find out value of the function at x equal to a f a is equal to cos a. Rarely we can see limit of the function is equal to value of the function at x equal to a. So we can write limit of x tending to a fx is equal to f a is equal to cos a. This implies function f is continuous at x equal to a or we can say function f is continuous at every real number. So we can write function f is continuous for all x belonging to r where r is the set of all real numbers or we can say cosine function is continuous for all x belonging to r. That is x belonging to real numbers. Let us now discuss the continuity of the cosecant function. Now let fx is equal to cosecant x. Now we can write it as fx equal to 1 upon sin x. We know cosecant function is reciprocal of sin function. So we can write cosecant x equal to 1 upon sin x. Now clearly we can see the given function f is not defined for x equal to pi 2 pi till n pi where n is any integer as for all these values denominator would be equal to 0. Now let us consider any real number a belonging to real numbers which do not include real numbers equal to n pi where n is any integer. Now let us find out limit of the function at x equal to a. So we can write limit of x tending to a fx is equal to limit of x tending to a 1 upon sin x. Now let us put x equal to a plus h then as x tends to a h tends to 0. Now this is equal to limit of h tending to 0 1 upon sin a plus h. Now this can be written as limit of h tending to 0 1 upon sin a cos h plus cos a sin h. We know sin a plus b is equal to sin a cos b plus cos a sin b. So here we have applied the formula for sin a plus b. Now this is equal to 1 upon sin a. Let us now find out value of the function at x equal to a that is f a f a is equal to 1 upon sin a. Clearly we can see limit of the function is equal to value of the function at x equal to a. So we can write limit of x tending to a fx is equal to f a is equal to 1 upon sin a. This implies function f is continuous at x equal to a. Or we can write cos e can function is continuous for all x belonging to r minus n pi where n belongs to integers. Let us now discuss the continuity of the secant function. So first of all let us assume that fx is equal to secant x. Now this can be written as fx equal to 1 upon cos x. Now clearly we can see function is not defined at these values as denominator will become 0 at these values. Now let us consider any real number a belonging to set of real numbers which do not include the real numbers equal to 2n plus 1 multiplied by pi upon 2 where n is any integer. Let us now find the limit of the function at x equal to a. So we can write limit of x tending to a fx is equal to limit of x tending to a 1 upon cos x which is equal to 1 upon cos a. Now let us find out value of the function at x equal to a. This is also equal to 1 upon cos a. Clearly we can see limit of the function is equal to value of the function. So this implies given function f is continuous at x equal to a. So we can write function f is continuous at x equal to a. Or we can say secant function is continuous for all x belonging to set of real numbers which do not include real numbers in equal to 2n plus 1 multiplied by pi upon 2 where n is any integer. Let us now discuss the continuity of cotangent function. First of all let us assume let fx equal to cot x. Or we can write fx equal to 1 upon tan x. We know cot x is reciprocal of tan x. Clearly we can see function is not defined at x equal to 0 pi 2 pi till n pi where n is any integer. As at these values of x denominator will be equal to 0. So function is not defined at these values. Now let us consider any real number a belonging to set of real numbers which do not include real numbers equal to n pi where n is any integer. Let us now find out limit of the function at x equal to a. We can write limit of x tending to a fx is equal to limit of x tending to a 1 upon tan x which is further equal to 1 upon tan a. Now let us find out value of the function at x equal to a. We get f a is equal to 1 upon tan a. Clearly we can see value of the function is equal to limit of the function at x equal to a. Now this implies function f is continuous at x equal to a. Or we can say function f is continuous at all x belonging to real numbers which do not include real numbers equal to n pi where n is any integer. So our required answer is cosine function is continuous for all x belonging to real numbers. Consequent function is continuous except for x equal to n pi where n belongs to integers. Second function is continuous except for x equal to 2n plus 1 multiplied by pi upon 2 where n belongs to integers. Consequent function is continuous except for x equal to n pi where n belongs to integers. This is our required answer. This completes the session. Hope you understood the session. Take care and goodbye.