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 ...

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= …

fact that #tanx# is an odd function (#tan (-x)=-tanx#) Identity #tanx=sinx/cosx# Answer link Ratnaker Mehta Sep 15, 2017

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?

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 ...

1/sqrt (2) tanx + cotx = 2 tanx + 1/tanx = 2 tan^2x/ (tanx) + 1/tanx = 2 (tan^2x + 1)/ (tanx) = 2 tan^2x + 1 = 2tanx tan^2x - 2tanx + 1 = 0 let a = tanx a^2 - 2a + 1 ...

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 …

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 ...