Given the graph, we want to find the equation for the tangent to the curve at P where P is π halves, one, so I found the one. Apparently I didn't need to do that if I'd looked at the graph. I went by the equation Y equal 3 plus cotangent X -, 2 cosec and XI stuck in Pi halves everywhere and I got Y of π halves equaling one which was given in the graph. So now we needed Y prime so we can find the slope of this line. So I take the derivative of the given equation and I get negative cosecant squared X + 2 cosecant X cotangent X sticking in Pi halves and knowing my unit circle negative cosecant squared Pi halves. Well, cosecant Pi halves is one and 1 * 1 is 1 + 2. Cosecant Pi halves is one. Again, cotangent Pi halves is 0, so our slope here is -1 using point slope form y -, 1 equal -1 times the quantity X - π halves. Distributing out the -1 and adding the -1 that was on the left side over to the right, we get Y equal negative X + π halves plus one. Now the second part says the horizontal tangent to the curve at Q. So if it's a horizontal tangent, we know the first derivative has to equal 0. So I'm going to take that first derivative, set it equal to 0 and a factor out of cosecant X. So I know that cosecant X = 0 and also negative cosecant X + 2, cotangent X = 0. Well, cosecant X is never zero. It has to be in between negative Infinity and -1 and one to Infinity. So cosec an X is never 0. So looking at the second one negative cosec an X + 2 cotangent X = 0. I'm going to add the cosec an X across and I realized that I get 2 cosec an X over sine X equaling one over sine X. So I'm going to multiply each side by sine X, making sure to recall in the end that sine X can't equal 0 if I get an angle that would have it as that. So we get 2 cosine X equal 1. Cosine X is 1/2. Knowing my unit circle, I know that cosine X is 1/2 at π thirds also at 5 Pi thirds. But when we look at the graph, we can see that Q is less than P&P was Pi halves. So we're wanting the π thirds location. So now to get the actual height, we're going to put that π thirds into the original equation. Y = 3 plus cotangent π thirds -2 cosecant π thirds, cotangent Π thirds by knowing our unit circle is root 3 / 3 cosecant π thirds. Once again knowing our unit circle is 2 root 3 / 3, so we get the three and root 3 / 3 - 4. Root 3 is over three, so 1 -, 4 is -3 Root 3 is over three or just negative root 3.