1. ## 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??

2. 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 $\displaystyle \cos{x}$ ...

3. ohh.. haha. thanks.. for some reason i never see the easiness of it .. ugh

4. 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

5. 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" ?

$\displaystyle \displaystyle\frac{tanx}{1+tanx}=\left[\frac{cosx}{cosx}\right]\frac{tanx}{1+tanx}$

6. 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.

7. 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?

8. 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.

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# tanx 1/tanx=1/ sinx cosx

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