Question

How do you find at equation of the tangent line to the graph `y=4+cotx-2cscx` at x=pi/2?

Answer 1

`y=-x+(pi/2+2)`

Explanation:

To find the equation of this tangent line we have to find the slope of this line at `x=pi/2` that is
`color(red)(slope = y' at x=pi/2`

`y_(pi/2)=4+cot(pi/2)-2csc(pi/2)`

`y_(pi/2)=4+0-2*1`
`y_(pi/2)=2`

this line passes through `(color(blue)(pi/2,2))`

To find the `color(red)(slope)` we should find `color(red)(y'_(pi/2))`

`y=4+cotx-2cscx`

`y'=4'+(cotx)'-2(cscx)'`

`y'=0-csc^2x-2(-cscxcotx)`
`y'=-csc^2x+2cscxcotx`

`color(red)(y'_(pi/2)=-csc^2(pi/2)+2csc(pi/2)cot(pi/2)`

`color(red)(y'_(pi/2)=-1+2*0=-1)`

The Equation of the tangent line with slope`-1` and passing through `(color(blue)(pi/2,2))`
:
`y-color(blue)(y_0)=color(red)(y'_(pi/2))(x-color(blue)(x_0))`
`y-2=-1(x-pi/2)`

`y=-x+(pi/2+2)`