# Pre-calculus

Prove the following identities.
1. 1+cosx/1-cosx = secx + 1/secx -1
2. (tanx + cotx)^2=sec^2x csc^2x
3. cos(x+y) cos(x-y)= cos^2x - sin^2y

1. 👍 0
2. 👎 0
3. 👁 620
1. correction for #1, should say:

(1+cosx)/(1-cosx) = (secx + 1)/(secx - 1)

RS = (1/cosx + 1)/(1/cosx - 1)
= (1/cosx + 1)/(1/cosx - 1) * cosx/cox
= (1+ cosx)/(1 - cosx)
= LS
Well, that was easy.

#2, hint: change everything to sines and cosines
expand and simplify the LS

#3, use the expansion for cos(A ± B), multiply the result and watch what happens.
hint: remember a^4 - b^4 = (a^2+b^2)(a^2-b^2)

1. 👍 0
2. 👎 0

## Similar Questions

1. ### math (trigonometry)

help me solve this coz im stumped secx-secx sin^2x= cosx thanks

Prove that cscx / cosx = tanx + cotx

secx/secx-tanx=sec^2x+ secx+tanx

4. ### Math

1)A piano tuner uses a tuning fork. If middle C has a frequency of 264 vibrations per second, write an equation in the form d=sinw(t) for the simple harmonic motion. 2) Verify the identity tan^2X-cot^2X/tanX+cotX=tanX-cotX I'm not

1. ### maths

proof that tanx + secx -1/tanx - secx+1 = 1+sinx/cosx

2. ### Trigonometry

Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do.

3. ### Trig Identities

Prove the following identities: 13. tan(x) + sec(x) = (cos(x)) / (1-sin(x)) *Sorry for any confusing parenthesis.* My work: I simplified the left side to a. ((sinx) / (cosx)) + (1 / cosx) , then b. (sinx + 1) / cosx = (cos(x)) /

4. ### precal

1/tanx-secx+ 1/tanx+secx=-2tanx so this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx) =sinx/cosx+ sinx /cosx= -2tanxI but I know this can't be correct because what I did doesn't end as a

Which of the following are identities? Check all that apply. (Points : 2) sin2x = 1 - cos2x sin2x - cos2x = 1 tan2x = 1 + sec2x cot2x = csc2x - 1 Question 4. 4. Which of the following equations are identities? Check all that