Question

What is the slope of the line normal to the tangent line of `f(x) = tanx-secx` at `x= (5pi)/12`?

Answer 1

Slope of the normal to the tangent line is `-1.457 (3dp)`

Explanation:

Slope of the tangent line at `x` is `dy/dx or f^'(x)`

`f (x) = tanx -secx` differentiating `f(x)` we get

`f^'(x) = sec^2 x - tanx*secx ; x= (5pi)/12 ~~ 1.309 (3dp)`

`f^'(1.309) = sec^2(1.309) - tan(1.309)*sec (1.309)`

`f^'(1.309) ~~ 0.50866 :.`. Slope of the tangent at `x=1.309`

is `m_t=0.50866` , since we know that the product of slopes of

two perpendicular lines is `-1 :.` Slope of the normal to the

tangent line is `m_n= -1/m_t = -1/0.50866 ~~ -1.457 (3dp)` [Ans]