Question

How do you find the equation of a line tangent to the function `y=-2tanx` at (-pi,0)?

Answer 1

The equation of the line is `y=-2(x+pi)`

Explanation:

`y=-2tanx`
`dy/dx=-2/cos^2x`
At the point `(-pi,0)` ; `y_0=0` and `dy/dx=-2`
Therefore the equqtion of the line is
`y-y_0=-2(x-x_0)`
`y=-2(x+pi)`
graph{(y+2tanx)(y+2x+2pi)=0 [-11.55, 2.5, -3.06, 3.96]}