Find an equation for the tangent to the curve at p and the horizontal tangent to the curve at q. y = 5 + cot x - 2 csc x 0 1 2 3 0 2 4 x y upper qupper p left parenthesis startfraction pi over 2 endfraction comma 3 right parenthesis

Answer with explanation:

The given function in x and y is,

  y= 5 +cot x-2 Cosec x

To find the equation of tangent, we will differentiate the function with respect to x

Slope of tangent at (π/2,3)

Equation of tangent passing through (π/2,3) can be obtained by

⇒There will be no Horizontal tangent from the point (π/2,3).

(4,5

look up a corrdinate plane and do what the thingy says and boom

i'm not sure i quite understand, could you provide more information?