Verify: (cotx / 1 - tanx) + (tanx / 1 - cotx) = 1 + tanx + cotx

Answer 1

It will suffice to show that the expression on the left side

MINUS the expression on the right side is zero...

recall, if A-B = 0 then A=B, where A is the left side

and B is the right side..

Also, The common denominator is (1-tan)(1-cot) =

1 - cot - tan + 1 =

2 - cot - tan <--- will subsitute

cot/(1 - tan) + tan/(1-cot) - [1 + tan + cot] =

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

cot (1 - cot) + tan(1-tan) - (1-cot)(1-tan) - tan(1-cot)(1-tan) - cot(1-cot)(1-tan)

=

cot - cot^2 + tan - tan^2 - (2 - cot - tan) - tan(2 - cot - tan) - cot(2 - cot - tan)

=

cot - cot^2 + tan - tan^2 -2 + cot + tan - 2 tan + 1 + tan^2 - 2cot + cot^2 + 1

=

0 <--- everything cancles

Answer 2

let x = 60 degrees, then cotx = 1/sqr3 and tanx = sqr3

(1/sqr3)/(1-sqr3) + sqr3/(1- 1/sqr3) = 1 + sqr3 + 1/sqr3

left side <0, right side >0

one counterexample disproves an "identity"

try x = pi/4 tanx=cotx = 1

left side is undefined,right side=3

this isn't an identity

Answer 3

 (cotx / 1 - tanx) + (tanx / 1 - cotx) = 1 + tanx + cotx

(1/sinx)/1-tanx + tanx -(1/sinx) = 1+tanx+1/sinx

sinx - " + " - " = " " "

sinx - " = " " "

" = " " +2/sinx ?????????