Question

What is the equation of the line tangent to `f(x)=-(sin2x)/sin^2 x` at `x=pi/4`?

Answer 1

Equation of tangent is `y=4x-(pi-2)`
Or `7y=28x-8`

Explanation:

`f(x)=(-sin2x)/sin^2x=(-2sinxcosx)/sin^2x=-2cotx`
`f’(x)=(-2)*(csc^2x)=2csc^2x`

For `x=pi/4`,
Slope of the tangent is
`f’(pi/4)=2csc^2(pi/4)=2(sqrt2)^2=4`

For `x=pi/4`, `y=f(pi/4)=-2cot(pi/4)=-2`

Equation of tangent is
`(y-y1)=m*(x-x1)`
`(y+2)=4(x-(pi/4))`
`y+2=4x-pi`