1. ## 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

2. ## 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.

3. ## Re: Proving an identity

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

4. ## Re: Proving an identity

Wait, i think I get it. Hold on.

5. ## 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?

Correct!