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

##### 1 Answer
Sep 21, 2017

Slope of the normal to the tangent line is $- 1.457 \left(3 \mathrm{dp}\right)$

#### Explanation:

Slope of the tangent line at $x$ is $\frac{\mathrm{dy}}{\mathrm{dx}} \mathmr{and} {f}^{'} \left(x\right)$

$f \left(x\right) = \tan x - \sec x$ differentiating $f \left(x\right)$ we get

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

${f}^{'} \left(1.309\right) = {\sec}^{2} \left(1.309\right) - \tan \left(1.309\right) \cdot \sec \left(1.309\right)$

${f}^{'} \left(1.309\right) \approx 0.50866 \therefore$. 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 $- 1 \therefore$ Slope of the normal to the

tangent line is ${m}_{n} = - \frac{1}{m} _ t = - \frac{1}{0.50866} \approx - 1.457 \left(3 \mathrm{dp}\right)$ [Ans]