Question

Prove the trig identity by working with only one side of the equation:

Answer 1

\[\frac{ \cos x * \sin x }{ \cos x - \sin x } = \frac{ \cos x }{ 1-\tan x } - \cos x\]

Answer 2

I have been trying to solve this for a few hours now... the hard part for me is that I am only supposed to work with ONE side of the equation. Any suggestions?

Answer 3

RHS
\[\large\rm \frac{ cosx }{ 1-tanx} -cosx \]
common denominator
\[\large\rm= \frac{ cosx -cosx(1-tanx) }{ 1-tanx } \rightarrow \frac{ cosx -cosx +cosx tanx }{ 1-tanx }\]
\[\large\rm = \frac{ cosx -cosx +cosx tanx }{ 1-tanx }\]
rewrite tanx as sinx/cosx
\[\large\rm = \frac{ cosx -cosx +cosx (\frac{sinx}{cosx}) }{ 1-\frac{sinx}{cosx} }\]

Answer 4

\[\large\rm = \frac{ \cancel{cosx -cosx} +\cancel{cosx} (\frac{sinx}{\cancel{cosx}}) }{ 1-\frac{sinx}{cosx} }\]
\[\frac{ sinx }{ 1-\frac{sinx}{cosx} }\]
simplify

Answer 5

opps \[\large\rm = \frac{ cosx -cosx +cosx (\frac{sinx}{cosx}) }{ 1-\frac{sinx}{cosx} }\]
\[\large\rm =\frac{ sinx }{ 1-tanx }\] multiply the numerator and denominator by the conjugate of `1-tanx` and then simplify

Answer 6

ok let me try that... 1 sec

Answer 7

take ur time :=))

Answer 8

\[\frac{ \sin x (1 + \tan x) }{ 1-\tan ^{2}x }\]

Answer 9

rewrite tanx as sin/cos \[\large\rm =\frac{ sinx +\frac{\sin^2x}{cosx} }{ 1-\frac{\sin^2x}{\cos^2x}} \]
common denominator
\[\large\rm =\frac{ \frac{sinx \cdot cosx +{\sin^2x}}{cosx} }{\frac{ cos^2x-\sin^2x}{\cos^2x}} \]
simplify

Answer 10

how did you get

\[\frac{ \sin^2x }{ \cos x }\]

from

\[1+tanx\]

Answer 11

good question i'm glad you askd that
the numerator is sinx(1+tanx)
rewrite tanx as sinx/cosx \[sinx(1+\frac{sinx}{Cosx}) \rightarrow sinx + \frac{\sin^2x}{cosx}\]

Answer 12

I get:
\[sinx (1+ \frac{ sinx }{ cosx })\]

Answer 13

yes and then distribute by sinx

Answer 14

haha... oops

Answer 15

tunnel vision

Answer 16

:=))

Answer 17

I'm at: \[\frac{ sinx * cosx + \sin^2x }{ cosx - \sin^2x }\]

Answer 18

can you explain how you got that..
hmm i'm confused

Answer 19

i missed some squares... one sec... need to rework

Answer 20

ok

Answer 21

\[\frac{ sinx*cosx+\sin^2x*cosx }{ \cos^2x-sinx }\]

Answer 22

i need to distribute that cosx in the numerator now i think..

Answer 23

Im thing about divide both top and bottom by cos(x), assuming cos(x) non zero, then we have sin over (1-tan), and the rest is reverse of your first and second post

Answer 24

yes correct i guess you forgot to square sinx

Answer 25

and this formula holds trivially if cos(x) is zero and that take care of the exception

Answer 26

i got:

\[\frac{ cosx + sinx }{ cosx - sinx }\]

not sure where I went wrong this time...

Answer 27

i have to go in few mintues let me post the solution \[\huge\rm =\frac{\frac{ sinx \cdot cosx+\sin^2x }{ cosx }}{\frac{\cos^2x-\sin^2x}{cos^2x}}\]\[\huge\rm =\frac{ sinx \cdot cosx+\sin^2x }{ cosx } \cdot \frac{cos^2x}{cos^2x-sin^2x}\]
\[\large \rm =\frac{ sinx \cdot cosx+\sin^2x }{ cosx } \cdot \frac{cosx \cdot cosx}{cos^2x-sin^2x}\]
\[\large \rm =\frac{ sinx \cdot cosx+\sin^2x }{\cancel{ cosx} } \cdot \frac{\cancel{cosx} \cdot cosx}{cos^2x-sin^2x}\]

\[\large \rm =\frac{ sinx \cdot cosx+\sin^2x }{ 1 } \cdot \frac{ cosx}{cos^2x-sin^2x}\]
\[\large \rm =\frac{ (sinx \cdot cosx+sin^2x )\cdot cosx }{ cos^2x-sin^2x }\]
rewrite it
\[\large \rm =\frac{ cosx(sinx \cdot cosx+sin^2x ) }{ cos^2x-sin^2x }\]
factor out sinx
\[\large \rm =\frac{ cosx \cdot sinx(1 \cdot cosx+sinx ) }{ cos^2x-sin^2x }\]
use the fact a^2-b^2=(a+b)(a-b)

Answer 28

i see what I did. i didnt factor our the sinx and I canceled out incorrectly

Answer 29

Thank you so much! You are the best!

Answer 30

my pleasure :=)) o^_^o