Verifying the following identity

• January 26th 2010, 09:19 PM
iamanoobatmath
Verifying the following identity
Okay hai guys so I'm trying to verify the following identity: $\frac{\sin^2\theta+2\cos\theta-1}{\sin^2\theta+3\cos\theta-3}=\frac{1}{1-\sec\theta}$

My attempt:

$\frac{1-\cos^2\theta+2\cos\theta-1}{1-cos^2\theta+3\cos\theta-3}$

$\frac{-\cos^2\theta+2\cos\theta}{-cos^2\theta+3\cos\theta-2}$

$\frac{-\cos\theta(\cos\theta-2)}{(\cos\theta-1)(\cos\theta-2)}$

$-\frac{\cos\theta}{\cos\theta-1}$

Now I'm going to work on the right side

$\frac{1}{1-\sec\theta}$

$\frac{1}{1-\frac{1}{\cos\theta}}$

$\frac{1}{\frac{\cos\theta-1}{cosx}}$

$\frac{cosx}{\cos\theta-1}$

So $-\frac{\cos\theta}{\cos\theta-1}=\frac{cosx}{\cos\theta-1}$

:confused: What am I doing wrong?
• January 26th 2010, 11:50 PM
Prove It

$\frac{-\cos^2{\theta} + 2\cos{\theta}}{-\cos^2{\theta} + 3\cos{\theta} - 2} = \frac{-\cos{\theta}(\cos{\theta} - 2)}{-\cos^2{\theta} + 2\cos{\theta} + \cos{\theta} - 2}$

$= \frac{-\cos{\theta}(\cos{\theta} - 2)}{-\cos{\theta}(\cos{\theta} - 2) + 1(\cos{\theta} - 2)}$

$= \frac{-\cos{\theta}(\cos{\theta} - 2)}{(-\cos{\theta} + 1)(\cos{\theta} - 2)}$

$= \frac{-\cos{\theta}}{-\cos{\theta} + 1}$

$= \frac{\cos{\theta}}{\cos{\theta} - 1}$
• January 26th 2010, 11:51 PM
Quote:

Originally Posted by iamanoobatmath
Okay hai guys so I'm trying to verify the following identity: $\frac{\sin^2\theta+2\cos\theta-1}{\sin^2\theta+3\cos\theta-3}=\frac{1}{1-\sec\theta}$

My attempt:

$\frac{1-\cos^2\theta+2\cos\theta-1}{1-cos^2\theta+3\cos\theta-3}$

$\frac{-\cos^2\theta+2\cos\theta}{-cos^2\theta+3\cos\theta-2}$

$\frac{-\cos\theta(\cos\theta-2)}{(\cos\theta-1)(\cos\theta-2)}$

$-\frac{\cos\theta}{\cos\theta-1}$ ***

Now I'm going to work on the right side

$\frac{1}{1-\sec\theta}$

$\frac{1}{1-\frac{1}{\cos\theta}}$

$\frac{1}{\frac{\cos\theta-1}{cosx}}$

$\frac{cosx}{\cos\theta-1}$

So $-\frac{\cos\theta}{\cos\theta-1}=\frac{cosx}{\cos\theta-1}$

:confused: What am I doing wrong?

hi

your factorisation part is wrong ,

continuing from ***, where the negative sign shouldn't be there,

$\frac{\cos \theta}{\cos \theta-1}$

$\frac{\frac{1}{\sec \theta}}{\frac{1}{\sec \theta}-1}$

Then combine the fractions is the denominator , and see where you can go from there .