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

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Answer 1 · 2017-12-03T12:46:51

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

Slope of the tangent is $f'(x)or f'((5pi)/6)$

$f(x)=tanx + sin(x-pi/4)$

$:.f'(x)=sec^2x + cos(x-pi/4)$

$:.f'((5pi)/6)=sec^2((5pi)/6) + cos((5pi)/6-pi/4)$

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

$:.f'((5pi)/6)=sec^2(150) + cos(150-45)$ or

$:.f'((5pi)/6)=1.33 -0.26 ~~1.07 or m_t ~~ 1.07$

Slope of the tangent line is $m_t ~~ 1.07$

Slope of normal to the tangent line is $m_n=-1/m_t$ or

$m_n= -1/1.07 ~~ -0.93$

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

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