# What is the slope of the line normal to the tangent line of #f(x) = tanx+sin(x-pi/4) # at # x= (5pi)/8 #?

##### 1 Answer

Dec 29, 2016

#### Explanation:

At #x=5/8pi,

y= f(5/8pi)=tan(5/8pi)+sin(5/8pi-pi/4)#

The slope of the normal at this point

is -1/y/ at P

The equation to the normal at P(1.9635,2.0315)# is

y-2.0315=-0.1290(x-1.9635)#. Simplifying,