Please help me with these calculus problems: establishing the identities eeek?

Question

you can answer all or one
please write out all the steps taken :-)

#1 (sin^3+cos^3)/(1-2cos^2)=
(sec-sin)/(tan-1)

#2 (tanA+tanB)(1-cotAcotB)+
(cotA+cotB)(1-tanAtanB)=0

#3 cos((3pi/2)+angle)=sin angle

#4( cos(A+B))/(cos(A-B))=
(1-tanAtanB)/(1+tanAtanB)

#5 sec(A-B)= (secAsecB)/(1+tanAtanB)

#6 (cos(A-B))(cos(A+B))= ((cos^2)A)-((sin^2)B)

A= alpha B=beta
thank you :-)

Answer 1

Breaking your individual terms up into sums of products of cos A, sin A, cos B, and sin B, then using trig identities to combine them into the form on the other side often helps. Let's take #5 as an example:

sec (A-B)
= 1/(cos (A-B)) (Rule: sec a = 1/cos a)
= 1/(cos A cos B + sin A sin B) (Rule: cosine difference formula)
= [1/(cos A cos B)] / [(cos A cos B + sin A sin B) / (cos A cos B)] (The fact that you need tan A tan B in the denominator on the right, coupled with the fact that you have a sin A sin B term, plus the identities tan a = sin a / cos a and sec a = 1 / cos a gives you a clue that you need to divide top and bottom by cos A cos B)
= [1/(cos A cos B)] / [(cos A cos B) / (cos A cos B) + (sin A sin B) / (cos A cos B)] (Rule: distributing the common term in the denominator
= [1/(cos A cos B)] / (1 + tan A tan B) (Simplifying the denominator)
= sec A sec B / (1 + tan A tan B) (Using the identity sec a = 1/ cos a in the numerator)

Answer 2

somewhere in your book you will find formulas for
cos(A + B) and cos(A - B)

For number 3, just let A = 3pi/2 (whose sine and cosine you know!) and let B= angle

For number 4, write out the formulas for numerator and denominator and then divide top and bottom by the cosines
(since tan = sin/cos)