How can I factor the algebraic expression 1 – csc²(x) in terms of a single trigonometric function?

To factor the expression 1 – csc²(x) in terms of a single trigonometric function, we first need to remember the relationship between the cosecant function and the sine function. The cosecant function is defined as:

csc(x) = 1/sin(x)

Therefore, we can write:

csc²(x) = 1/sin²(x)

Substituting this into our expression gives:

1 – csc²(x) = 1 – (1/sin²(x))

To simplify this expression, we need a common denominator:

1 can be rewritten as sin²(x)/sin²(x). Now the expression becomes:

1 – csc²(x) = (sin²(x)/sin²(x)) – (1/sin²(x))

Now, we have:

1 – csc²(x) = (sin²(x) – 1) / sin²(x)

We can factor the numerator, sin²(x) – 1, which is a difference of squares:

sin²(x) – 1 = (sin(x) – 1)(sin(x) + 1)

This leads us to:

1 – csc²(x) = ((sin(x) – 1)(sin(x) + 1)) / sin²(x)

Thus, we have factored the algebraic expression in terms of sine, a single trigonometric function:

1 – csc²(x) = -cot²(x)

So the final factored form is:

1 – csc²(x) = -cot²(x)

This result is useful not only for simplifying calculations but also for solving problems involving trigonometric identities.

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