1)
In general, given a function f(x), f'(a) is the slope of the tangent line to the graph of f(x) at x=a.
Thus, in our case, calculating the derivative of y(x),
[tex]\begin{gathered} y=5+cot(x)-2csc(x) \\ \Rightarrow y^{\prime}(x)=-csc^2x-2(-cot(x)csc(x))=-csc^2x+2cot(x)csc(x) \end{gathered}[/tex]
Then, evaluate y'(x) at the x-coordinate of point P (x=pi/2), as shown below
[tex]y^{\prime}(\frac{\pi}{2})=-(1)^2+2*0*1=-1[/tex]
Therefore, the slope of the tangent line to the curve at P is equal to -1.
Calculating the equation of such line given its slope and a point on it,
[tex]\begin{gathered} P=(\frac{\pi}{2},3),m=-1 \\ \Rightarrow y-3=-1(x-\frac{\pi}{2}) \\ \Rightarrow y=-x+\frac{\pi}{2}+3= \end{gathered}[/tex]
The answer to the first part is y=-x+(6+pi)/2, which is equivalent to y=-x+pi/2+3
(pi/2+3=4.570796...)
2)According to the question, the line is the horizontal tangent to the curve at Q; therefore, its slope is equal to zero because it is horizontal.
Then, we need to find the y-coordinate of point Q. Notice that the x-coordinate of point Q is 1; thus,
[tex]y(1)=5+cot(1)-2csc(1)\approx3.2653...[/tex]
Therefore, the exact equation is y=5+cot(1)-2csc(1), or approximately y=3.2653...
