1. ## 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,

2. 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:

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

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

3. 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:

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

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

Distribute the minus sign to the denominator:

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

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

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

5. 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,

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

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

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

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

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

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

$=1-\sin^2 B$

$=\cos^2 B$