# What is the slope of the line normal to the tangent line of f(x) = tanx+sin(x-pi/4)  at  x= (5pi)/6 ?

Dec 3, 2017

Slope of the normal to the tangent line is $- 0.93$

#### Explanation:

Slope of the tangent is $f ' \left(x\right) \mathmr{and} f ' \left(\frac{5 \pi}{6}\right)$

$f \left(x\right) = \tan x + \sin \left(x - \frac{\pi}{4}\right)$

$\therefore f ' \left(x\right) = {\sec}^{2} x + \cos \left(x - \frac{\pi}{4}\right)$

$\therefore f ' \left(\frac{5 \pi}{6}\right) = {\sec}^{2} \left(\frac{5 \pi}{6}\right) + \cos \left(\frac{5 \pi}{6} - \frac{\pi}{4}\right)$

(5pi)/6=150^0 ; pi/4=45^0

$\therefore f ' \left(\frac{5 \pi}{6}\right) = {\sec}^{2} \left(150\right) + \cos \left(150 - 45\right)$ or

$\therefore f ' \left(\frac{5 \pi}{6}\right) = 1.33 - 0.26 \approx 1.07 \mathmr{and} {m}_{t} \approx 1.07$

Slope of the tangent line is ${m}_{t} \approx 1.07$

Slope of normal to the tangent line is ${m}_{n} = - \frac{1}{m} _ t$ or

${m}_{n} = - \frac{1}{1.07} \approx - 0.93$

Slope of the normal to the tangent line is $- 0.93$ [Ans]