# Proving an identity

Printable View

• Jul 17th 2011, 11:52 AM
raymac62
Proving an identity
Having some trouble proving this identity.

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

This is where I keep getting stuck.

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

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

Help please.

Thanks in advance.

Sincerely,

Raymond
• Jul 17th 2011, 11:56 AM
Bruno J.
Re: Proving an identity
$\displaystyle \cos x + \frac{\sin^2 x}{\cos x} \neq \frac{1+\sin^2x}{\cos x}$. You're not putting these under a common denominator properly.
• Jul 17th 2011, 12:04 PM
raymac62
Re: Proving an identity
Quote:

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

Is it...
$\displaystyle \cos x + \frac{\sin^2 x}{\cos x} = \frac{cosx+\sin^2x}{\cos x}$ ?
• Jul 17th 2011, 12:06 PM
raymac62
Re: Proving an identity
Wait, i think I get it. Hold on.
• Jul 17th 2011, 12:13 PM
raymac62
Re: Proving an identity
$\displaystyle cosx+ sinxtanx= \frac {1}{cosx}$

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

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

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

Correct now?
• Jul 17th 2011, 12:23 PM
Siron
Re: Proving an identity
Correct!