00:01
So in this question we're asked to verify the identity relationship.
00:05
And the first step would be to choose the more complex side, nor to manipulate.
00:08
And in this case, it would be the left -hand side, as we have more trig functions within it, and we have a new brain denominator that could easily simplify and reduce, so that we can eventually equal the other half of the relationship, 1 minus sine squared x.
00:23
So our first step would be just to simplify the equation.
00:27
And we can do that by first, simplifying kotan x, which is equal to cosine x, over sine x all over cosine x over sine x over sine x plus sine x over cosine we can further simplify this by actually applying a fraction rule where if we have say terms b c over a can eventually rearrange the terms in order equal b over c times a in this case, b would be the numerator of the numerator, which is cosine x.
01:12
C would be sine x, and a would be the whole denominator.
01:17
So essentially we can rewrite this by saying cosine x over sine x times the denominator.
01:26
Cosine x over sine x plus sine x over cosine.
01:37
And what can make this simpler is instead of of having the two fractions separate from each other we can actually combine them into a common denominator and then we would have cosine x over sine x times we would need sine x times cosine x as the common denominator so we'd apply the opposite we would multiply by the other term that is not present already in each terms denominator so basically in the first especially we have cosine x times cosine x.
02:17
We have sine x already in the numerator.
02:19
So it's cosine squared x over plus and for the same thing with the opposite other terms, sine x has cosine x as cosine x is denominator, so we'll multiply by sine x.
02:30
We have sine squared x all over sine x cosine x and from here we would move on by finally multiplying sine x through and when we do that, we see that there's already a sinex in the denominator.
02:57
So we can actually cancel that out.
02:59
And we're left with cosine x over cosine squared x plus sine squared x over sine squared x over cosine x.
03:13
And from there we can apply the rule where we're essentially dividing by the reciprocal...