Hello! I have been learning about proving identities and can do simple ones but lately I have been really struggling. I am currently stuck on this problem.

cos(A-B)/tan(A)+cot(B) = sin(B)/sec(A)

I managed to get to:
cosAcosB-sinAsinB/(sinA/cosA) + (cosA/sinB) but I am stuck

5 answers

hint: cos(A-B) = cosAcosB + sinAsinB

tan(A)+cot(B)
= sinA/cosA + cosB/sinB
= (sinAsinB + cosAcosB)/(cosAsinB) , using a common denominator.

now, what happens when you perform the division?
Hello! Thank you for responding, I did some cancelling and now I have

cosAcosB-sinAsinB/sinA+cosB

If I cancel the cosB I end up with

cosA-sinAsinB/sinA

am I allowed to cancel the sinA if the one on top is a negative while the one at the bottom is positive?
Oh sorry I just noticed it is + not -!
I ended up with cosA+sinB :(
LS = (cosAcosB + sinAsinB) / ((sinAsinB + cosAcosB)/(cosAsinB) ), from my first post
= (cosAcosB + sinAsinB)(cosAsinB) / (sinAsinB + cosAcosB)
= cosAsinB
= sinB (1/secA)
= RS
Similar Questions
    1. answers icon 1 answer
  1. Trigonometry identities are so hard...I need some help proving these identities: *Oh, and I'm only in grade 11, so the
    1. answers icon 5 answers
  2. Proving Trigonometric Identities1. sec^2x + csc^2x= (sec^2 x)(csc^2 x) 2. sin ^3 x / sin x - cos 3x / cos x = 2 3. 1- cos x/ sin
    1. answers icon 1 answer
    1. answers icon 0 answers
more similar questions