Explanation: #1/cos^2x=1-tanx# Identities: #color (red) (1/cos^2x=sec^2x)# #color (red) (sec^2x=1+tan^2x)# So we have: #1+tan^2x=1-tanx# Subtract #1# from both sides: #tan^2x=-tanx# …

LHS= (1-sec x)/tan x - tan x/ (1 - sec x) = (1-sec x)/tan x + (tan x (secx-1))/ ( ( sec x-1) (secx+1)) =1/tanx-sec x/tan x + (tan x (secx-1))/ ( sec^2 x-1) =1/tanx ...

How do you differentiate f (x) = x tan 3x + x3 tan x using the product rule?

Kindly refer to the Explanation. "Prerequisites : " (1) : 1+cos2theta=2cos^2theta, : (2) : 1-cos2theta=2sin^2theta. cot^2 (pi/4+x/2)= (cos^2 (pi/4+x/2))/sin^2 (pi/4+x ...

Apr 5, 2018 · How would you prove the following equation? # (secx)/ (1-tanx) = (1)/ (cosx-sinx)# Thank y'all for the help!

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 ...

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 words, the …

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 ...

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 …

Explanation: Identities: #color (red) (1+cot^2x=csc^2x)# #color (red) (csc^2x=1/sin^2x# #color (red) (cotx=1/tanx#