Proving an identity

**raymac62** · July 17th 2011, 11:52 AM

Having some trouble proving this identity.

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

This is where I keep getting stuck.

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

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

Help please.

Thanks in advance.

Sincerely,

Raymond

**Bruno J.** · July 17th 2011, 11:56 AM

$\cos x + \frac{\sin^2 x}{\cos x} \neq \frac{1+\sin^2x}{\cos x}$ . You're not putting these under a common denominator properly.

**raymac62** · July 17th 2011, 12:04 PM

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...
$\cos x + \frac{\sin^2 x}{\cos x} = \frac{cosx+\sin^2x}{\cos x}$ ?

**raymac62** · July 17th 2011, 12:06 PM

Wait, i think I get it. Hold on.

**raymac62** · July 17th 2011, 12:13 PM

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

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

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

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

Correct now?

**Siron** · July 17th 2011, 12:23 PM

Correct!

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