Step-by-step explanation:
We have
1-cotĀ²a + cotā“a = sinĀ²a(1+cotā¶a)
First, we can take a look at the right side. It expands to sinĀ²a + cotā¶(a)sinĀ²(a) = sinĀ²a + cosā¶a/sinā“a (this is the expanded right side) as cot(a) = cos(a)/sin(a), so cosā¶a = cosā¶a/sinā¶a. Therefore, it might be helpful to put everything in terms of sine and cosine to solve this.
We know 1 = sinĀ²a+cosĀ²a and cot(a) = cos(a)/sin(a), so we have
1-cotĀ²a + cotā“a = sinĀ²a+cosĀ²a-cosĀ²a/sinĀ²a + cosā“a/sinā“a
Next, we know that in the expanded right side, we have sinĀ²a + something. We can use that to isolate the sinĀ²a. The rest of the expanded right side has a denominator of sinā“a, so we can make everything else have that denominator.
sinĀ²a+cosĀ²a-cosĀ²a/sinĀ²a + cosā“a/sinā“a
= sinĀ²a + (cosĀ²(a)sinā“(a) - cosĀ²(a)sinĀ²(a) + cosā“a)/sinā“a
We can then factor cosĀ²a out of the numerator
sinĀ²a + (cosĀ²(a)sinā“(a) - cosĀ²(a)sinĀ²(a) + cosā“a)/sinā“a
= sinĀ²a + cosĀ²a (sinā“a-sinĀ²a+cosĀ²a)/sinā“a
Then, in the expanded right side, we can notice that the fraction has a numerator with only cos in it. We can therefore write sinā“a in terms of cos (we don't want to write the sinĀ²a term in terms of cos because it can easily add with cosĀ²a to become 1, so we can hold that off for later) , so
sinĀ²a = (1-cosĀ²a)
sinā“a = (1-cosĀ²a)Ā² = cosā“a - 2cosĀ²a + 1
sinĀ²a + cosĀ²a (sinā“a-sinĀ²a+cosĀ²a)/sinā“a
= sinĀ²a + cosĀ²a (cosā“a-2cosĀ²a+1-sinĀ²a+cosĀ²a)/sinā“a
= sinĀ²a + cosĀ²a (cosā“a-cosĀ²a+1-sinĀ²a)/sinā“a
factor our the -cosĀ²a-sinĀ²a as -1(cosĀ²a+sinĀ²a) = -1(1) = -1
sinĀ²a + cosĀ²a (cosā“a-cosĀ²a+1-sinĀ²a)/sinā“a
= Ā sinĀ²a + cosĀ²a (cosā“a-1 + 1)/sinā“a
= sinĀ²a + cosā¶a/sinā“a
= sinĀ²a(1+cosā¶a/sinā¶a)
= sinĀ²a(1+cotā¶a)
