i have a few questions

Question

cosx + cosx =2secx
----- -----
1+sinx 1-sinx

cos(x-B)-cos(x-B) = 2sinxsinB

csc2x= 1 secx cscx
---
2

cotx= cos5x+cos3x
-----------
sin 5x-sin 3x

Answer 1

for the first one:

LS = cosx/(1+sinx) + cosx/(1-sinx)

getting a common denominator of (1+sinx)(1-sinx)
= [cosx(1-sinx) + cosx(1+sinx)]/(1+sinx)(1-sinx)
= 2cosx/(1-sin^2)
= 2cosx/cos^2x
= 2/cosx
= 2secx
= RS

for cos(x-B)-cos(x-B) = 2sinxsinB

use the formula cos(A-B) = cosAcosB + sinAsinB on the Left Side, it comes apart very nicely after that

For the last two, try changing all trig ratios into sines and cosines.
Show me what you get

Answer 2

for cos(x-B)-cos(x-B) = 2sinxsinB
you must have a typo, the LS is zero the way it stands
I am sure you meant

cos(x-B)-cos(x+B) = 2sinxsinB

Answer 3

i don't understand how to use the formula in the second problem. what would i plug into A and B?

Answer 4

cos(x-B)-cos(x+B) = 2sinxsinB

LS
= cosxcosB + sinxsinB - (cosxcosB - sinxsinB)
=2sinxsinB
= RS