1-tan x / 1+tan x = cot x -1 / cot x +1
2007-01-04 04:53:27 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Divide the top and the bottom of the left hand side by tan x.
2007-01-04 04:58:25 · answer #1 · answered by Phineas Bogg 6 · 0⤊ 0⤋
You should write the equation to be proved in a standard way such as,
(1-tan x) /( 1+tan x) = (cot x -1) / (cot x +1)
Proof.
(1-tan x) /( 1+tan x)
=[(1-tan x)cot x] /[( 1+tan x)cot x]
= (cot x - 1)/(cot x +1)
END
2007-01-04 13:04:22 · answer #2 · answered by sahsjing 7 · 2⤊ 0⤋
devide both numerator n denominator by tan x
whats the prob i dont get it
2007-01-04 12:59:41 · answer #3 · answered by well thts it...... 3 · 0⤊ 0⤋
(1-sin x/cos x)/(1+sin x/cos x)
cos x/sin x((1-sin x/cos x)/(1+sin x/
cos x))
((cos x/sin x)-1)/((cos x/sin x)+1)
cot x-1/cot x+1
2007-01-04 13:15:20 · answer #4 · answered by Anonymous · 3⤊ 0⤋
I'm guessing you need to prove the equation.Sorry,can't help you there.
2007-01-04 13:01:05 · answer #5 · answered by Cheng J 2 · 0⤊ 1⤋
lsft hand side
LHS
=[1-sinx/cosx]/[1+sinx/cosx]
=[cosx-sinx]/[cosx+sinx]....1
RHS
=[cosx/sinx-1]/[cosx/sinx+1]
=[cosx-sinx]/[cosx+sinx].....2
we observe from 1.&2.that both LHS and
RHS are equating to same quantity.
Hence proved
2007-01-04 13:05:54 · answer #6 · answered by openpsychy 6 · 0⤊ 0⤋
cotx=1/tanx
so
RHS= (1/tanx -1 ) /(1/tanx +1)
=[(1-tanx)/tanx ] / [ (1+tanx)/tanx]
==[(1-tanx)/tanx ] / [tanx/ (1+tanx)]
=(1-tanx)/(1+tanx)
=LHS
2007-01-04 13:22:22 · answer #7 · answered by iyiogrenci 6 · 1⤊ 0⤋