How do I reduce and simplify this problem?

Question

sec^4 B - csc^4 B / sec^2 - csc^2 B

I'm not sure on how to do this problem.

Answer 1

Remember that, by definition,

sec B = 1 / cos B, and csc (or cosec) B = 1 / sin B.

Making those substitutions as soon as possible often simplifies the task, since certain things to do with sines and cosines are generally much more familiar.

WAIT A MINUTE! DID YOU NEGLECT TO PUT BRACKETS IN (as well as one of the B's)? You really MUST learn to be more careful and specific --- those of us trying to help you are NOT clairvoyant!

This looks such an unholy b*****d without brackets, I'm going to insert them anyway, and assume that the question is REALLY about:

(sec^4 B - csc^4 B) / (sec^2 B - csc^2 B).

Don't you SEE how very different THAT is?!

O.K. You said that you're not sure how to do it, but I've just found that my suggestion, above, held the key.

Separately work out the numerator and denominator, in the sine and cosine language. Each of them will itself need to be expressed as a fraction. In one of them, some factorization of one term and elimination of one of the factors (via the most well known and quoted trig. identities, hint, hint) will simplify it slightly.

When you then divide one of these fractions by the other, it should all simplify down to:

1/(cos^2 B sin^2 B) = sec^2 B csc^2 B, back in the original sec, csc language.

I took the form of your question to mean that you would like to be put onto the right path, without being spoonfed throughout with the full solution.

Now go to it. Good luck!

Live long and prosper.

P.S. I confess that in trying to unravel what you really meant, I missed the fact that the numerator in the correct way of writing your expression could be factorized as the difference of two squares. Using this, the corrected expression is of course:

sec^2 B + csc^2 B = 1 /( cos^2 B sin^2 B) = sec^2 B csc^2 B.

(Note once again: the substitution originally suggested, and the mental use of the world's best known trig. identity, sin^2 B + cos^2 B = 1 --- in the second stage immediately above --- quickly complete the final stage of the simplification.)

Answer 2

(sec^2 B + csc^2 B)(sec^2 B - csc^2 B)/(sec^2 B - csc^2 B)

you're left with
sec^2 B + csc^2 B

that is
1/(cos^2 B) + 1/(sin^2 B)
the common denom is sin^2 B cos^2 B
so we have (sin^2 B + cos^2 B)/(sin^2 B cos^2 B)
that's csc^2 B sec^2 B

Answer 3

if you will notice the numerator is a difference of two squares namely (sec^2B)^2-(csc^2B)^2 factoring it will give the identity
(sec^2B+csc^2B)(sec^2B-csc^2B)

as you can see one of the factor of the numerator is the same as denominatos so we can cancel that one already making your equation

sec^2B+csc^2B
which i think you can simplify further

Answer 4

sec^4B-csc^4B=(sec^2B+csc^2B)(sec^2B-csc^2B)
so sec^4b-csc^4B/sec^2b-csc^2B=sec^2B+csc^2B
=1+tan^2B+1+cot^2B
=2+tan^2B+cot^2B
=tan^2B+cot^2B+2tanBcotB (tanBcotB=1)
=(tanB+cotB)^2