# trig id

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• Apr 17th 2009, 01:51 PM
pberardi
trig id
Hey all, I was wondering if someone can walk me through this one:

[cot^2(B) - cos^2(B)]/[csc^2(B) -1]

My solution is as follows:
[cot^2(B)/-sec^2(B)] - cos^2(B)/-sec^2(B)

-cos^4(B)/sin^2(B) +1

and now i am a bit stuck. the correct answer is cos^2(B). Can someone please finish this off or tell me a better way to get to the right answer?

Thanks,
• Apr 17th 2009, 03:19 PM
e^(i*pi)
Quote:

Originally Posted by pberardi
Hey all, I was wondering if someone can walk me through this one:

[cot^2(B) - cos^2(B)]/[csc^2(B) -1]

My solution is as follows:
[cot^2(B)/-sec^2(B)] - cos^2(B)/-sec^2(B)

-cos^4(B)/sin^2(B) +1

and now i am a bit stuck. the correct answer is cos^2(B). Can someone please finish this off or tell me a better way to get to the right answer?

Thanks,

From where you got to:

Distribute the minus sign to the denominator:

$\displaystyle \frac{cos^4(B)}{-(sin^2(B)+1)} = \frac{cos^4(B)}{1-sin^2(B)}$

and what does 1-sin^2(B) equal (Wink)
• Apr 17th 2009, 03:31 PM
JoanF
Quote:

Originally Posted by pberardi
Hey all, I was wondering if someone can walk me through this one:

[cot^2(B) - cos^2(B)]/[csc^2(B) -1]

My solution is as follows:
[cot^2(B)/-sec^2(B)] - cos^2(B)/-sec^2(B)

-cos^4(B)/sin^2(B) +1

and now i am a bit stuck. the correct answer is cos^2(B). Can someone please finish this off or tell me a better way to get to the right answer?

Thanks,

Hi!

I think you did wrong on the signs. Because you should get -cos^4(B)/sin^2(B) - 1. If you get this, you get the exercice correct:

http://www.postimage.org/Pq2gAaNS.jpg

If you want to see my resolution, here it is:

http://www.postimage.org/aV2fHtqi.jpg
• Apr 17th 2009, 04:15 PM
JoanF
Quote:

Originally Posted by e^(i*pi)
From where you got to:

Distribute the minus sign to the denominator:

$\displaystyle \frac{cos^4(B)}{-(sin^2(B)+1)} = \frac{cos^4(B)}{1-sin^2(B)}$

and what does 1-sin^2(B) equal (Wink)

But
$\displaystyle \frac{cos^4(B)}{-(sin^2(B)+1)} =\frac{cos^4(B)}{-sin^2(B)-1}$
• Apr 17th 2009, 06:15 PM
Shyam
Quote:

Originally Posted by pberardi
Hey all, I was wondering if someone can walk me through this one:

[cot^2(B) - cos^2(B)]/[csc^2(B) -1]

My solution is as follows:
[cot^2(B)/-sec^2(B)] - cos^2(B)/-sec^2(B)

-cos^4(B)/sin^2(B) +1

and now i am a bit stuck. the correct answer is cos^2(B). Can someone please finish this off or tell me a better way to get to the right answer?

Thanks,

Here is a better way, see,

$\displaystyle \frac{\cot^2 B-\cos^2 B}{\csc^2 B-1}$

$\displaystyle =\frac{\cot^2 B-\cos^2 B}{\cot^2 B}$

$\displaystyle =\frac{\cot^2 B}{\cot^2 B}-\frac{\cos^2 B}{\cot^2 B}$

$\displaystyle =1-\frac{\cos^2 B}{\cot^2 B}$

$\displaystyle =1-\frac{\cos^2 B}{\frac{\cos^2 B}{\sin^2 B}}$

$\displaystyle =1-\cos^2 B\times \frac{\sin^2 B}{\cos^2 B}$

$\displaystyle =1-\sin^2 B$

$\displaystyle =\cos^2 B$