# Thread: Establish the Identity ((1+cosx+sinx)/(1+cosx-sinx)) = secx + tanx

1. ## Establish the Identity ((1+cosx+sinx)/(1+cosx-sinx)) = secx + tanx

Hi!

I need to establish this: using the more basic trigonometric identities.

I've tried all kinds of stuff for several hours without results. Any help would be highly appreciated!

2. ## Re: Establish the Identity ((1+cosx+sinx)/(1+cosx-sinx)) = secx + tanx

Originally Posted by Benciticus
Hi!

I need to establish this: using the more basic trigonometric identities.

I've tried all kinds of stuff for several hours without results. Any help would be highly appreciated!
maybe there is an easier way ???

$\displaystyle \frac{(1+\cos{x})+\sin{x}}{(1+\cos{x})-\sin{x}} \cdot \frac{(1+\cos{x})+\sin{x}}{(1+\cos{x})+\sin{x}}$

$\displaystyle \frac{(1+\cos{x})^2 + 2(1+\cos{x})\sin{x} + \sin^2{x}}{(1+\cos{x})^2 - \sin^2{x}}$

$\displaystyle \frac{1+2\cos{x}+\cos^2{x}+2\sin{x}+2\cos{x}\sin{x }+\sin^2{x}}{1+2\cos{x}+\cos^2{x}-\sin^2{x}}$

$\displaystyle \frac{2+2\cos{x}+2\sin{x}+2\cos{x}\sin{x}}{2\cos{x }+2\cos^2{x}}$

$\displaystyle \frac{2(1+\cos{x})+2\sin{x}(1+\cos{x})}{2\cos{x}(1 +\cos{x})}$

$\displaystyle \frac{2(1+\cos{x})[1+\sin{x}]}{2(1+\cos{x})\cos{x}}$

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

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

$\displaystyle \sec{x} + \tan{x}$

... what a pain-in-the-@ identity!

3. ## Re: Establish the Identity ((1+cosx+sinx)/(1+cosx-sinx)) = secx + tanx

Awesomesauce!!

Thanks a bunch!

4. ## Re: Establish the Identity ((1+cosx+sinx)/(1+cosx-sinx)) = secx + tanx

Alternatively write the right hand side as 1/cosx+sinx/cosx= (1+sinx)/cosx and then show that LHS-RHS=0 by putting on common denominator.

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# (1 cosx-sinx)/(1-cosx-sĩn)

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