178k views
4 votes
Prove that the expression a⋅cot(x)−b⋅csc(x), where a>b, has at least one value of c for which f(c) is defined, given that f:R→R.

User Jon Cahill
by
7.7k points

1 Answer

2 votes

Final answer:

To prove that the expression a⋅cot(x)−b⋅csc(x) has at least one value of c for which f(c) is defined, we need to consider values of x for which sin(x) is not zero.

Step-by-step explanation:

To prove that the expression a⋅cot(x)−b⋅csc(x) has at least one value of c for which f(c) is defined, we can start by understanding the definitions of cotangent and cosecant.

The cotangent of an angle x is defined as the ratio of the adjacent side to the opposite side of a right triangle formed by x. The cosecant of an angle x is defined as the reciprocal of the sine of x, which is equal to 1 divided by the sine of x.

Since the expression involves cot(x) and csc(x), we need to make sure that the angles x we consider are such that the cotangent and cosecant are defined. This means that the sine of x should not be zero, since the sine of an angle is the reciprocal of the cosecant. Therefore, x cannot be any angle for which sin(x) = 0.

In conclusion, as long as we consider values of x for which sin(x) is not zero, the expression a⋅cot(x)−b⋅csc(x) will have at least one value of c for which f(c) is defined.

User Milind Dumbare
by
8.2k points