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# stuck on a few trig questions watch

1. hey,
can you help me on these questions
prove identities of following:

cot A/1 + cot A ≡ 1/1 + tan A

sec A - cos A ≡ tan A sin A

sin A/1 - cos A ≡ cosec A + cot A

thanks
2. For the second one write the tan on the RHS as sin/cos and then apply the pythagorean identity.

Similar with the others, start by writing everything in terms of sin/cos and it is easier to see what to do.
3. for the 2nd part i do this

1/cos - cos ≡ sin/cos x sin

what would i do next ?
4. ?
5. how u get that ?
6. (Original post by Lehmber Hussainpuri)
how u get that ?
It's the pythagorean identity, apply it to the RHS of your expression and job done.
7. (Original post by Lehmber Hussainpuri)
hey,
can you help me on these questions
prove identities of following:

cot A/1 + cot A ≡ 1/1 + tan A

sec A - cos A ≡ tan A sin A

sin A/1 - cos A ≡ cosec A + cot A

thanks
Sada lembher bulleh bulleh bulleh lol well you got to familiarise yourself with the trig identities then manipulate them to prove those above questions for example question the working out is:

[spoiler]
sec A = 1/cos A therefore the occasion sec A - cos A is same as 1/cos A - cos A which is the same as (1 - cos^2 A)/cos A using the trig identiy sin^2 A + cos^2 A =1 you manipulate to get sin^2 A/cos A which is same tan A sin A
[\spoiler]
8. Also, start with one side, and then try and manipulate that into the other side, don't be changing them both at the same time.

These really come with practice, but it is worth it because it will pay dividends by the time you are integrating trig functions.
9. (Original post by Lehmber Hussainpuri)
hey,
can you help me on these questions
prove identities of following:

cot A/1 + cot A = 1/1 + tan A

thanks
Have you not heard of brackets ?

cot A/(1 + cot A) = 1/(1 + tan A)
10. with

cot A/(1 + cot A) = 1/(1 + tan A)

tackle the LHS

change (1 + cot A) to (1 + tan A)/tan A {from the fact that cot A = 1/tan A}
11. ok thanks, i get it now.

can you help me on another question:

State the period, in radians, of the graph y = cosec 4x

thanks
LEHMBER
12. Do you understand the question? What does it mean by period? If you know, try drawing the graph to help you.
13. I can draw the graph, but don't understand the period bit?
14. the period bit just means how long does the graph take to repeat
15. (Original post by Lehmber Hussainpuri)
I can draw the graph, but don't understand the period bit?
Ugh. So rather than "can you help me with this question", what you actually meant was "can you explain the period of a function", right? Please, we're not mindreaders.

The period of a function f is the smallest number a such that f(x) = f(x+a) for all x. So the period of sin is 2pi, since sin x = sin(x+2pi) for all x. As the poster above me said, this is more or less the amount you go along the x-axis before the graph starts to repeat itself.

If f(x) = cosec 4x, you're looking for the (smallest) number a such that cosec 4x = cosec 4(x+a).
16. woo i get it now THANKS,

I am stuck on a few other questions;

given that 2cos^2x - Sin^2x = 1 show tht cos^2x = 2 sin^2x

given that 5sec^2x + 3 tan^2x = 9

solve the equation cot^2 + cosecx = 11, all solutions between 0<x<360

thnks
17. (Original post by steve2005)
Have you not heard of brackets ?

cot A/(1 + cot A) = 1/(1 + tan A)
lol i wondered what the question was saying

I am stuck on a few other questions;

given that 2cos^2x - Sin^2x = 1 show tht cos^2x = 2 sin^2x

given that 5sec^2x + 3 tan^2x = 9

solve the equation cot^2 + cosecx = 11, all solutions between 0<x<360

thnks
the first one

the second one;

And with the last thing
you need to factorise with completing the square.
18. (Original post by Lehmber Hussainpuri)
woo i get it now THANKS,

I am stuck on a few other questions;

given that 2cos^2x - Sin^2x = 1 show tht cos^2x = 2 sin^2x

given that 5sec^2x + 3 tan^2x = 9

solve the equation cot^2 + cosecx = 11, all solutions between 0<x<360

thnks
Care to show any working...?

They all involve the identity cos²x+sin²x=1 and the two related identities.

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