Question

How we do this this is hoighs chbool clac 1 i failed it and i have to reatek it

Answer 1

The equation of the curve is given by:

`y=5+\cot(x)-2\csc(x)`

Differentiating both sides of the equation with respect to x, we have:

`(dy)/(dx)=2\cot(x)\csc(x)-\csc^2(x)`

Therefore, the slope of the tangent is given by the value of dy/dx when x= π / 2

`2\cot((\pi)/(2))\csc((\pi)/(2))-\csc^2((\pi)/(2))=-1`

Using the point slope formula, it follows that:

`\begin{gathered} y-3=-1(x-(\pi)/(2)) \\ y=-x+(\pi)/(2)+3 \end{gathered}`

Therefore, the equation of the tangent at P is given by:

y = -x + π /2 + 3