lim_ (x->0) (1-cosx)/tan^2x = 1/2 Using the trigonometric identities: 2sin^2 (x/2) = 1-cosx and: tan x = (2tan (x/2))/ (1-tan^2 (x/2)) we have: (1-cosx)/tan^2x= (2sin ...

Apr 29, 2017 · Here is my proof, which is wrong? I am proving sinx - sinxtanx =0 sinx = sinxtanx 1 = tanx Apparently I HAVE to factor out sinx. Why is that? Trigonometry

Mar 14, 2018 · sinx(1 −2sin2x +sin4x)sec5(x) = tanx We can now factor the parenthesis: sinx(1 −sin2x)(1 − sin2x)sec5(x) = tanx A very familiar modified Pythagorean identity will now be …

Given tan (cotx)=cot (tanx) => tan (cotx)=tan (pi/2-tanx) =>cotx+tanx=pi/2 =>cosx/sinx+sinx/cosx=pi/2 => (cos^2x+sin^2x)/ (sinxcosx)=pi/2 =2/ (2sinxcosx)=pi/2 =>2 ...

If we factor the equation: sinx (tanx-1)=0 So: sinx=0rArrx=kpi and tanx-1=0rArrtanx=1rArrx=pi/4+kpi

Nov 18, 2017 · Then, EE" a "c in (a,b)" such that "f' (c)=0. In our case, f (x)=tanx, a=0, b=pi. Note that, f (x)=tanx, is not defined at x=pi/2 in [0,pi]. Hence, f is not continuous on (o,pi). In other …

Mar 30, 2018 · Explanation: As #y=cos (sinx)# # (dy)/ (dx)=-sin (sinx)*cosx# i.e. #sin (sinx)=-1/cosx (dy)/ (dx)# and using product formula # (d^2y)/ (dx^2)=-cosxcos (sinx)cosx+sin ...

d/ (dx) (1/ (secx-tanx))=secxtanx+sec^2x As sec^2x=tan^2x+1, we have sec^2x-tan^2x=1 i.e. (secx+tanx) (secx-tanx)=1 and 1/ (secx-tanx)=secx+tanx Hence d/ (dx) (1 ...

How do you find the derivative of [sec x(tan x + cos x)]?

Equation of normal line is x+ 1.17y =0.87 x=pi/8 ~~0.39 ; f (x)=tanx = or f (x)= tan (pi/8) ~~ 0.41 So at (0.39,0.41) tangent and normal is drawn. Slope of the tangent is f' (x)= sec^2x. at x=pi/8 …