Question

1) 1+cos(3t)/ sin(3t) + sin(3t)/( 1+ cos(3t))= 2csc(3t)

2) sec^2 2u-1/ sec^2 2u= sin^2 2u

3) cosB/1- sinB= secB+ tanB

Answer 1

Please type your equations using brackets to show the proper order of operation.

The way you typed them, none are identities.

e.g.
for the last one you probably meant:
cosB/(1-sinB) = secB + tanB

then:
RS = 1/cosB + sinB/cosB
= = (1-sinB)/cosB * (1+sinB)/(1+sinB)
= (1 - sin^2 B)/(cosB(1+sinB)
= cos^2B/(cosB(1+sinB))
= cosB/(1+sinB)
= LS

Answer 2

1)1+cos(3t)/ sin(3t) + sin(3t)/( 1+ cos(3t))= 2csc(3t)




2)sec^2 (2u-1)/ sec^2 (2u)= sin^2 (2u)

Answer 3

in 1) since all angles are 3t, I will let x = 3t for easier typing

Again, the way you typed it, it does not work.
You must have meant
(1+cosx)/sinx + sinx/(1+cosx) = 2cscx

LS = ((1+cosx)^2 + sin^2x)/(sinx(1+cosx))
= (1 + 2cosx + cos^2x + sin^2x)/(sinx(1+cosx))
= (2+2cosx)/(sinx(1+cosx))
= 2(1+cosx)/(sinx(1+cosx))
= 2/sinx
= 2cscx, but x = 3t
= 2csc(3t)
= RS

2) does not work the way you typed it.

Answer 4

(sin^2 è+ cos^2 è)^3 =1

Answer 5

(sin^2 è+ cos^2 è)^3 =1

That is much too easy!!!!

what is the value of sin^2 è+ cos^2 è ?