Help with trigonometry again. Verifying identities. cot x+1/cot x-1 =1+tan x/ 1- tanx?

Question

I need a 95% on next test to keep 'A' in class help to understand

Answer 1

Consider the left-hand side(LHS) [cot(x) + 1] / [cot(x) – 1]. We want to prove that this is equal to the right-hand side (RHS), that is, [1 + tan(x)] / [1 – tan(x)]. Thus, we have the following:

LHS = [cot(x) + 1] / [cot(x) + 1]
        = { [cot(x) + 1] / [cot(x) – 1] }* { tan(x) / tan(x) }.

NOTE: We multiply tan(x) to cot(x) + 1 and the other tan(x) is multiplied to cot(x) – 1.

        = { tan(x) [cot(x) + 1] } / { tan(x) [cot(x) – 1] }

        = { tan(x) cot(x) + tan(x)] } / { tan(x) cot(x) – tan(x)] }, by distributive property of multiplication over addition (subtraction)

        = { 1 + tan(x)] } / { 1 – tan(x)] }, since tan(x) cot(x) = 1 by reciprocal identity.

        = RHS.

Indeed, [cot(x) + 1] / [cot(x) – 1] = [1 + tan(x)] / [1 – tan(x)]

Answer 2

It is elementary as cot is in LHS and tan on RHS

convert cot to 1/tan

LHS = (cot x +1)/(cot x-1) = (1/tan x +1)/(1/tan x -1)

= ((1+tan x)/tan x)/((1-tan x)/tan x)
= (1+tan x)/(1-tan x) multiply numerator and denominator by tan x

Answer 3

tan x/(a million-cot x) = tanx(a million+cot x) /(a million-cot^2 x) = (a million+tan x)/(a million-cot^2x) tan x/(a million-cot x) = tanx(a million-cot x) /(a million-cot^2 x) = (-a million+tan x)/(a million-cot^2x) upload the two to get 2 tan x/(a million-cot^2 x) = 2 tan^3x/(tan^2 x -a million) = - tan x tan 2x RHS = a million + sec x cosec x = a million + a million/cos x a million/sin x = (a million+ cos x sin x)/ cos x sin x

Answer 4

cot x+1/cot x-1 =? 1+tan x/ 1- tanx
(1 - tanx)(1 + cotx) =? (1 + tanx)(cotx - 1)
1 + cotx - tanx -1 =?cotx - 1 + 1 - tanx
cotx - tanx = cotx - tanx

Answer 5

U take LHS or RHS, get the other, just use cotx = 1/tanx or
tanx = 1/cotx,,, whichever side u are comfortable with,, just algebra (common denom) cracks...

Answer 6

just divide (cotx+1)/(cotx-1) through by
cotx and you have (1+tanx)/(1-tanx)

it's that easy

i hope that this helps