Simplify the expression
(sin^2(t)-2sin(t)+1)/(sin(t)-1)
( a - b ) ^ 2 = a ^ 2 - 2 a * b + b ^ 2
[ sin ( t ) - 1 ] ^ 2 =
sin ^ 2 ( t ) - 2 * sin ( t ) * 1 + 1 ^ 2 =
sin ^ 2 ( t ) - 2 sin ( t ) + 1
so :
sin ^ 2 ( t ) - 2 sin ( t ) + 1 = [ sin ( t ) - 1 ] ^ 2
[ sin ^ 2 ( t ) - 2 sin ( t ) + 1 ] / [ sin ( t ) - 1 ] =
[ sin ( t ) - 1 ] ^ 2 / [ sin ( t ) - 1 ] =
sin ( t ) - 1
[ sin ^ 2 ( t ) - 2 sin ( t ) + 1 ] / [ sin ( t ) - 1 ] = sin ( t ) - 1
I don't understand why you squared (sin(t)-1)
To simplify the expression, we can factor the numerator and denominator.
Let's start by factoring the numerator.
The numerator is a quadratic expression in terms of sin(t). We can rewrite it as (sin(t))^2 - 2sin(t) + 1.
Now, notice that this is a perfect square trinomial, which can be factored as (sin(t) - 1)^2.
So, the numerator can be simplified to (sin(t) - 1)^2.
Now, let's move on to factoring the denominator.
The denominator is sin(t) - 1, which does not factor further.
Therefore, the simplified expression is (sin(t) - 1)^2 / (sin(t) - 1).
However, we can simplify this even further.
Since both the numerator and denominator have the factor (sin(t) - 1), they cancel each other out.
Therefore, the simplified expression is just sin(t) - 1.