Simplity (sec^(2)x-1)/(sec^(2)x-1)

Identify Expression: Identify the expression to be simplified. We are given the expression (sec^(2)x-1)/(sec^(2)x-1) and asked to simplify it.Recognize Identical Terms: Recognize that the numerator and denominator are identical. The expression (sec^(2)x-1)/(sec^(2)x-1) has the same terms in both the numerator and the denominator.Apply Property: Apply the property that any non-zero quantity divided by itself equals 1. Since the numerator and denominator are the same and assuming sec^(2)x-1 is not equal to zero, the expression simplifies to 1.Consider Undefined Condition: Consider the condition where the expression is undefined. The expression sec^(2)x-1 is equal to zero when sec^(2)x = 1. This happens when x is any integer multiple of π, because sec(x) = 1/cos(x) and cos(πn) = (-1)^n, which equals 1 for even n and -1 for odd n. However, sec(x) cannot be 1 when cos(x) is -1, so we only consider even multiples of π. Therefore, the original expression is undefined for x = πn where n is an even integer.

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Q. Simplity

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\frac{\sec^{2}x-1}{\sec^{2}x-1}

Identify Expression: Identify the expression to be simplified. $\newline$ We are given the expression $(\sec^{2}x-1)/(\sec^{2}x-1)$ and asked to simplify it.
Recognize Identical Terms: Recognize that the numerator and denominator are identical. The expression $(\sec^{2}x-1)/(\sec^{2}x-1)$ has the same terms in both the numerator and the denominator.
Apply Property: Apply the property that any non-zero quantity divided by itself equals $1$ . Since the numerator and denominator are the same and assuming $\sec^{2}x-1 \neq 0$ , the expression simplifies to $1$ .
Consider Undefined Condition: Consider the condition where the expression is undefined. $\newline$ The expression $\sec^{2}x-1$ is equal to zero when $\sec^{2}x = 1$ . This happens when $x$ is any integer multiple of $\pi$ , because $\sec(x) = \frac{1}{\cos(x)}$ and $\cos(\pi n) = (-1)^n$ , which equals $1$ for even $n$ and $-1$ for odd $n$ . However, $\sec^{2}x = 1$ $0$ cannot be $1$ when $\sec^{2}x = 1$ $2$ is $-1$ , so we only consider even multiples of $\pi$ . Therefore, the original expression is undefined for $\sec^{2}x = 1$ $5$ where $n$ is an even integer.