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
[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).
