Asked by Jenny
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
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
Answers
Answered by
Reiny
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?
tan(A)+cot(B)
= sinA/cosA + cosB/sinB
= (sinAsinB + cosAcosB)/(cosAsinB) , using a common denominator.
now, what happens when you perform the division?
Answered by
Jenny
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?
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?
Answered by
Jenny
Oh sorry I just noticed it is + not -!
Answered by
Jenny
I ended up with cosA+sinB :(
Answered by
Reiny
LS = (cosAcosB + sinAsinB) / ((sinAsinB + cosAcosB)/(cosAsinB) ), from my first post
= (cosAcosB + sinAsinB)(cosAsinB) / (sinAsinB + cosAcosB)
= cosAsinB
= sinB (1/secA)
= RS
= (cosAcosB + sinAsinB)(cosAsinB) / (sinAsinB + cosAcosB)
= cosAsinB
= sinB (1/secA)
= RS
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