1-cot^2a+cot^4a=sin^2a(1+cot^6a) prove it.​

Step-by-step explanation:

We have

1-cot²a + cot⁴a = sin²a(1+cot⁶a)

First, we can take a look at the right side. It expands to sin²a + cot⁶(a)sin²(a) = sin²a + cos⁶a/sin⁴a (this is the expanded right side) as cot(a) = cos(a)/sin(a), so cos⁶a = cos⁶a/sin⁶a. Therefore, it might be helpful to put everything in terms of sine and cosine to solve this.

We know 1 = sin²a+cos²a and cot(a) = cos(a)/sin(a), so we have

1-cot²a + cot⁴a = sin²a+cos²a-cos²a/sin²a + cos⁴a/sin⁴a

Next, we know that in the expanded right side, we have sin²a + something. We can use that to isolate the sin²a. The rest of the expanded right side has a denominator of sin⁴a, so we can make everything else have that denominator.

sin²a+cos²a-cos²a/sin²a + cos⁴a/sin⁴a

= sin²a + (cos²(a)sin⁴(a) - cos²(a)sin²(a) + cos⁴a)/sin⁴a

We can then factor cos²a out of the numerator

sin²a + (cos²(a)sin⁴(a) - cos²(a)sin²(a) + cos⁴a)/sin⁴a

= sin²a + cos²a (sin⁴a-sin²a+cos²a)/sin⁴a

Then, in the expanded right side, we can notice that the fraction has a numerator with only cos in it. We can therefore write sin⁴a in terms of cos (we don't want to write the sin²a term in terms of cos because it can easily add with cos²a to become 1, so we can hold that off for later) , so

sin²a = (1-cos²a)

sin⁴a = (1-cos²a)² = cos⁴a - 2cos²a + 1

sin²a + cos²a (sin⁴a-sin²a+cos²a)/sin⁴a

= sin²a + cos²a (cos⁴a-2cos²a+1-sin²a+cos²a)/sin⁴a

= sin²a + cos²a (cos⁴a-cos²a+1-sin²a)/sin⁴a

factor our the -cos²a-sin²a as -1(cos²a+sin²a) = -1(1) = -1

sin²a + cos²a (cos⁴a-cos²a+1-sin²a)/sin⁴a

=  sin²a + cos²a (cos⁴a-1 + 1)/sin⁴a

= sin²a + cos⁶a/sin⁴a

= sin²a(1+cos⁶a/sin⁶a)

= sin²a(1+cot⁶a)

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