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 …

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 ...

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

Explanation: We know that, #y=u*v,# then, # (dy)/ (dx)=u* (dv)/ (dx)+v* (du)/ (dx)# Taking, #u=-e^x+tanxand v=x^2-2x# we get # (du)/ (dx)=-e^x+sec^2xand (dv)/ (dx)=2x ...

Feb 18, 2018 · #sin^2x-tan^2x=sin^2x (1-1-tan^2x)=-sin^2xtan^2x#

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= …

3 Answers 1s2s2p Jan 9, 2018 #cscx-=1/sinx# #secx-=1/cosx# So, now we have: # (cosx+sinx) (1/cosx+1/sinx)# We can now expand this to get: # (cosx*1/cosx)+ (sinx*1 ...

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

Jun 16, 2015 · How do you convert all of the terms to sines and cosines and simplify to find the expression that completes the identity: tanx/secx?