C3 Trig Proof Help!!!

Hi, I'm struggling with 2 questions asking to prove trig identities.

1) Prove sec²x (x) cosec²x= 2 + cot²x + tan²x

I've started on the right hand side and wrote 2sin²x + 2cos²x + (cos²x)(/sin²x) + (sin²x)/(cos²x), and then split up these fractions to get 3/sin^2x + 3/cos^x = 3cosec^2x + 3sec^2x = cosec^2x +sec^2x (however, the proof says x not +), what have I done wrong? How does the times sign appear? Could someone please help

2) Prove sec⁴x - tan⁴x = 2sec²x - 1

As for this question, I have no idea where to start since I get nowhere near the correct proof, could someone give me a hand?

Thank you!!! :smile:

(edited 9 years ago)

Scroll to see replies

Reply 1

TeeEm

Original post by CleverGirl383

Hi, I'm struggling with 2 questions asking to prove trig identities.

1) Prove sec²x x cosec²x= 2 + cot²x + tan²

I've started on the right hand side and wrote 2sin²x + 2cos²x + (cos²x)(/sin²x) + (sin²x)/(cos²x), and then split up these fractions to get 3/sin^2x + 3/cos^x = 3cosec^2x + 3sec^2x = cosec^2x +sec^2x (however, the proof says x not +), what have I done wrong? How does the times sign appear? Could someone please help

2) Prove sec⁴x - tan⁴x = 2sec²x - 1

As for this question, I have no idea where to start since I get nowhere near the correct proof, could someone give me a hand?

Thank you!!! :smile:

1st
try starting from LHS

2nd
Can you see the difference of squares in the LHS?

Reply 2

Super199

Original post by TeeEm

1st
try starting from LHS

2nd
Can you see the difference of squares in the LHS?

Mm im also stuck with the first one. Ive done the second one. How should I start the first one? I tried (1+tan^2x)(1+cot^2x) but don't seem to be getting anywhere. Any hints on where to start?

Edit: never mind I didn't see the 2 at the beginning oops :tongue:

(edited 9 years ago)

Reply 3

TeeEm

Original post by Super199

Mm im also stuck with the first one. Ive done the second one. How should I start the first one? I tried (1+tan^2x)(1+cot^2x) but don't seem to be getting anywhere. Any hints on where to start?

multiply out and note cotx = 1/tanx

Reply 4

CleverGirl383

For Q1, I've tried using LHS and got up to 1/cos²x x 1/sin²x = 1/cos²xsin²x = sin²x + cos²x /cos²xsin²x, but I'm getting tan²x + cot²x, and not the +2... what have I done?

As for Q2, (sec²x + tan²x)(sec²x - tan²x) is the difference of squares here, but how does that help me?

Reply 5

TeeEm

Original post by CleverGirl383

look carefully at the second bracket

Reply 6

CleverGirl383

Original post by TeeEm

look carefully at the second bracket

I've wrote the second bracket as 1/cos^2x - sin^2x / cos^2x = 1-sin^2x / cos^2x = 1/cos^2x - sin^2x / cos^2x, getting an answer of sec^2x - 1. But I'm still somehow missing a 2 at the front of the sec term?

Reply 7

TeeEm

Original post by CleverGirl383

isnt the second bracket just 1? :smile:

Reply 8

CleverGirl383

Original post by TeeEm

isnt the second bracket just 1? :smile:

Yeah it is 1... So that leaves just (sec²x + tan²x)? do I have to set this to equal to 0? I tried to write it as 1/cos^2x + sin^2x / cos^2x and got 1+sin^2x / cos^2x, final answer was sec^2x + 1 (not the correct proof!), these are so confusing..

Reply 9

TeeEm

Original post by CleverGirl383

You do not set to 0 !!!!!

Use the obvious identity. You are line away from your answer!

Reply 10

CleverGirl383

Original post by TeeEm

You do not set to 0 !!!!!

Use the obvious identity. You are line away from your answer!

If I use an identity I can write 1 as sec²x - tan²x, so I end up with 2sec²x -tan²x (don't know where the -1 comes from?!)

Reply 11

TeeEm

Original post by CleverGirl383

If I use an identity I can write 1 as sec²x - tan²x, so I end up with 2sec²x -tan²x (don't know where the -1 comes from?!)

sec²x+tan²x = sec²x +(sec²x - 1)

Reply 12

CleverGirl383

Original post by TeeEm

sec²x+tan²x = sec²x +(sec²x - 1)

Is that an identity you're supposed to know? Never seen that before.

Reply 13

TeeEm

Original post by CleverGirl383

Is that an identity you're supposed to know? Never seen that before.

1+tan²x=sec²x

I believe the above identity is common knowledge but I could be wrong

Reply 14

CleverGirl383

Original post by TeeEm

1+tan²x=sec²x

I believe the above identity is common knowledge but I could be wrong

Ahh so that's how you derived it! Thanks very much, got the answer now for Q2! What about Question 1?

Reply 15

TeeEm

Original post by CleverGirl383

Ahh so that's how you derived it! Thanks very much, got the answer now for Q2! What about Question 1?

look at posts 2 and 4

Reply 16

CleverGirl383

Original post by TeeEm

look at posts 2 and 4

This is my method for Q1: Start LHS: 1/cos²x (x) 1/sin²x = sec²x + cosec²x --> (tan²x + 1) + (1 + cot²x) = 2 + cot²x + tan²x.

Is this correct?

Reply 17

TeeEm

Original post by CleverGirl383

This is my method for Q1: Start LHS: 1/cos²x (x) 1/sin²x = sec²x + cosec²x --> (tan²x + 1) + (1 + cot²x) = 2 + cot²x + tan²x.

Is this correct?

this is not correct

after the red you show that multiplication is addition

(edited 9 years ago)

Reply 18

CleverGirl383

Original post by TeeEm

this is not correct

after the red you show that multiplication is addition

Yes but whether you add or multiply, you get the exact same value, I did this using my calculator by substituting a number for x. Did you have another method?

Reply 19

TeeEm

Original post by CleverGirl383

Yes but whether you add or multiply, you get the exact same value, I did this using my calculator by substituting a number for x. Did you have another method?

sin0 + sin90 =sin(0+90) works

but sinA + sinB does not equal to the sin(A+B)

look at post 4

read what the other chap mentioned, then read my response.