Can you show mi working for this

Sundae · #1 $Old$ April 27th, 2009, 09:55 AM

(sec^4x - 2 sec^2x tan^2x + tan^4x) i need to simplify this equation and the answer is 1, i just dont know how 2 work it. Thx.

masters · #2 $Old$ April 27th, 2009, 10:01 AM

Quote:

Originally Posted by Sundae $View Post$

(sec^4x - 2 sec^2x tan^2x + tan^4x) i need to simplify this equation and the answer is 1, i just dont know how 2 work it. Thx.

Hi Sundae,

I assume the expression you have is set to 0.

$\sec^4x-2\sec^2 x \tan^2 x+ \tan^4x=0$

Now factor:

$(\sec^2 x-\tan^2 x)^2=0$

Edit: No solution to the above equation, anyhow. So, you just want to simplify the expression. See my last post.

Sundae · #3 $Old$ April 27th, 2009, 10:07 AM

its not equal to 0, i just have to simplify till i get to 1.

masters · #4 $Old$ April 27th, 2009, 10:17 AM

Quote:

Originally Posted by masters $View Post$

Hi Sundae,

I assume the expression you have is set to 0.

$\sec^4x-2\sec^2 x \tan^2 x+ \tan^4x=0$

Now factor:

$(\sec^2 x-\tan^2 x)^2=0$

Quote:

Originally Posted by Sundae $View Post$

its not equal to 0, i just have to simplify till i get to 1.

Ok sundae, now I understand.

$(\sec^2 x-\tan^2 x)^2$

$(\sec^2 x-\tan^2 x)(\sec^2 x-\tan^2 x)$

$\left(\frac{1}{\cos^2 x}-\frac{\sin^2 x}{\cos^2 x}\right)\left(\frac{1}{\cos^2 x}-\frac{\sin^2 x}{\cos^2 x}\right)$

$\left(\frac{1-\sin^2 x}{\cos^2 x}\right)\left(\frac{1-\sin^2 x}{\cos^2 x}\right)$

$\left(\frac{\cos ^2 x}{\cos^2 x}\right) \left(\frac{\cos ^2 x}{\cos^2 x}\right)=1$

Sundae · #5 $Old$ April 27th, 2009, 10:28 AM

Aw, that was really simple. Thankyou