Prove:(tanx/(1+tanx)) = (sinx/(sinx+cosx))

• Nov 15th 2010, 05:56 PM
tmas
Prove:(tanx/(1+tanx)) = (sinx/(sinx+cosx))
For this question the only answer I seem to get is sinx/cosx not (sinx/(sinx+cosx))

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

((1-tanx/1-tanx)) times (tanx/(1+tanx))

((tanx(1-tanx))/(1-tan^2x))

(tanx-tan^2x)/(1-tan^2x))

((cosx/cosx) times (sinx/cosx) - (sin^2x/cos^2x)) / ((cos^2x/cos^2x) times 1- (sin^2x/cos^2x))

((cosxsinx - sin^2x)/cos^2x) / ((cos^2x-sin^2x)/cos^2x)

multiply the reciprocal and cancel i get

sinx/cosx

ugh.. what did i do wrong??
• Nov 15th 2010, 06:11 PM
skeeter
you're really making this too hard ...

$\displaystyle \displaystyle \frac{\tan{x}}{1+\tan{x}} =$

$\displaystyle \displaystyle\frac{\frac{\sin{x}}{\cos{x}}}{1 + \frac{\sin{x}}{\cos{x}}}$

finish it by multiplying numerator and denominator by $\displaystyle \cos{x}$ ...
• Nov 15th 2010, 06:13 PM
tmas
ohh.. haha. thanks.. for some reason i never see the easiness of it .. ugh
• Feb 2nd 2011, 03:01 AM
ericpepin
One of the easiest way to prove this we can also use it as (1-tanx)/tanx= (sinx-cosx)/cosx. because we changing both numerator so this will be ok and both sides will give you result cotx-1 so this is proved. Thanks
• Feb 2nd 2011, 03:56 AM
Quote:

Originally Posted by tmas
For this question the only answer I seem to get is sinx/cosx not (sinx/(sinx+cosx))

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

((1-tanx/1-tanx)) times (tanx/(1+tanx))

((tanx(1-tanx))/(1-tan^2x))

(tanx-tan^2x)/(1-tan^2x))

((cosx/cosx) times (sinx/cosx) - (sin^2x/cos^2x)) / ((cos^2x/cos^2x) times 1- (sin^2x/cos^2x))

((cosxsinx - sin^2x)/cos^2x) / ((cos^2x-sin^2x)/cos^2x)

multiply the reciprocal and cancel i get

sinx/cosx

ugh.. what did i do wrong??

You went wrong on your final step.

$\displaystyle \displaystyle\frac{\left(cosx\;sinx-sin^2x\right)\frac{1}{cos^2x}}{\left(cos^2x-sin^2x\right)\frac{1}{cos^2x}}=\frac{cosx\;sinx-sin^2x}{cos^2x-sin^2x}$

$\displaystyle =\displaystyle\frac{sinx(cosx-sinx)}{(cosx-sinx)(cosx+sinx)}$

Simplest is to ask "how do we get from $\displaystyle tanx\rightarrow\ sinx$ in the numerator" ?

$\displaystyle \displaystyle\frac{tanx}{1+tanx}=\left[\frac{cosx}{cosx}\right]\frac{tanx}{1+tanx}$
• Feb 2nd 2011, 04:14 AM
DrSteve
Whenever you are trying to prove a trig identity with tan x, cot x, sec x or csc x, a method that usually works pretty easily is to change these into sin x and cos x and then simplify algebraically. So if you don't see a quicker way right away always try this.
• Feb 2nd 2011, 04:14 AM
skeeter
Quote:

Originally Posted by ericpepin
One of the easiest way to prove this we can also use it as (1-tanx)/tanx= (sinx-cosx)/cosx. because we changing both numerator so this will be ok and both sides will give you result cotx-1 so this is proved. Thanks

ok ... but what does that have to do with the original identity to be proved in this thread?
• Feb 2nd 2011, 08:46 PM
ericpepin
Quote:

Originally Posted by skeeter
ok ... but what does that have to do with the original identity to be proved in this thread?

I think here we trying to prove L.H.S.=R.H.S.