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 xTo find the equation of tangent, we will differentiate the function with respect to x[tex]y'= -\csc^2 x+2 \csc x\times \cot x[/tex]Slope of tangent at (π/2,3)  [tex]y'_{(\frac{\pi}{2},3)}= -\csc^2\frac{\pi}{2} +2 \csc \frac{\pi}{2}\times \cot \frac{\pi}{2}\\\\=-1+2\times 1 \times 0\\\\= -1[/tex]Equation of tangent passing through (π/2,3) can be obtained by [tex]\rightarrow \frac{y-y_{1}}{x-x_{1}}=m(\text{Slope})\\\\ \rightarrow \frac{y-3}{x-\frac{\pi}{2}}=-1\\\\\rightarrow 3-y=x-\frac{\pi}{2}\\\\\rightarrow x+y-3-\frac{\pi}{2}=0[/tex]⇒There will be no Horizontal tangent from the point (π/2,3).