# Thread: verifying identities help needed

1. ## verifying identities help needed

i have no idea! at all how to do it. plz help

1) Cos X/Sec X - 1 - Cos X/tan^2 X = Cot^2 X

2) Sin X / 1 - Cos X + 1 - Cos X / Sin X = 2CSC X

2. ## first identity

$\frac{\cos\theta}{\sec\theta-1}-\frac{\cos\theta}{\tan^2\theta}=\cot^2\theta$

$\frac{\cos\theta}{\frac{1}{\cos\theta}-1}-\frac{\cos\theta}{\frac{\sin^2\theta}{\cos^2\theta }}$

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

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

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

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

$\frac{\cos^2\theta}{\sin^2\theta}=\cot^2\theta$

3. ## second identity

$\frac{\sin\theta}{1-\cos\theta}+\frac{1-\cos\theta}{\sin\theta}=2\csc\theta$

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

$\frac{2\sin\theta}{\sin^2\theta}$

$\frac{2}{\sin\theta}=2\csc\theta$

4. Hello everyone
Originally Posted by satishgaire
...1) Cos X/Sec X - 1 - Cos X/tan^2 X = Cot^2 X...
Here's a little tip: where possible try to factorise an expression if you can, rather than expanding it. You'll usually keep the working much simpler that way.

Now if you don't know that
$\sec^2\theta = 1 + \tan^2\theta$
then you should learn it. (If you can't see where it comes from, just divide both sides of $\cos^2\theta+\sin^2\theta = 1$ by $\cos^2\theta$.)

Well, that means that:
$\tan^2\theta = \sec^2\theta - 1$
$=(\sec\theta -1)(\sec\theta +1)$...(1)
Can you see that this means that we can factorise the original expression? Like this:
$\frac{\cos\theta}{\sec\theta-1} -\frac{\cos\theta}{\tan^2\theta}=\frac{\cos\theta}{ \sec\theta-1} -\frac{\cos\theta}{(\sec\theta -1)(\sec\theta +1)}$, using (1)
$=\left(\frac{\cos\theta}{\sec\theta-1}\right)\left(1 -\frac{1}{\sec\theta +1}\right)$

$=\left(\frac{\cos\theta}{\sec\theta-1}\right)\left(\frac{\sec\theta}{\sec\theta +1}\right)$

$=\frac{1}{\tan^2\theta}$, using (1) again.

$=\cot^2\theta$