# Proving an identity

• Jul 17th 2011, 12:52 PM
raymac62
Proving an identity
Having some trouble proving this identity.

$cosx+ sinxtanx= \frac {1}{cosx}$

This is where I keep getting stuck.

$R.S=\frac {1}{cosx}$
$L.S= cosx+ sinx\frac {sinx}{cosx}$
$=cosx + \frac {sin^2x}{cosx}$

$=\frac {1 + sin^2x}{cosx}$

Sincerely,

Raymond
• Jul 17th 2011, 12:56 PM
Bruno J.
Re: Proving an identity
$\cos x + \frac{\sin^2 x}{\cos x} \neq \frac{1+\sin^2x}{\cos x}$. You're not putting these under a common denominator properly.
• Jul 17th 2011, 01:04 PM
raymac62
Re: Proving an identity
Quote:

Originally Posted by Bruno J.
$\cos x + \frac{\sin^2 x}{\cos x} \neq \frac{+\sin^2x}{\cos x}$. You're not putting these under a common denominator properly.

Is it...
$\cos x + \frac{\sin^2 x}{\cos x} = \frac{cosx+\sin^2x}{\cos x}$ ?
• Jul 17th 2011, 01:06 PM
raymac62
Re: Proving an identity
Wait, i think I get it. Hold on.
• Jul 17th 2011, 01:13 PM
raymac62
Re: Proving an identity
$cosx+ sinxtanx= \frac {1}{cosx}$

$R.S=\frac {1}{cosx}$

$L.S=\frac {cos x}{1} + \frac{\sin^2 x}{\cos x} = \frac{cos^2x}{cos x}+\frac {sin^2x}{cosx} = \frac {sin^2x+cos^2x}{cosx}$

$= \frac {1}{cosx}; R.S=L.S$

Correct now?
• Jul 17th 2011, 01:23 PM
Siron
Re: Proving an identity
Correct!