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

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Answer 1 · 2017-09-21T12:45:30

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

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]