P2 Question (Trig)

cot^2x - tan^2x = 4cot2x cosec2x

proof LHS = RHS

can anyone do this?

TheWolf

cot^2x - tan^2x = 4cot2x cosec2x

proof LHS = RHS

can anyone do this?

cot2x = (1 - tan^2 x)/2tanx
cosec2x = 1/(2sinx cosx)
LHS = 4cot2x cosec2x = (1-tan^2 x)/(sinx cosx tanx) = (1 - tan^2 x)/sin^2 x = cosec^2 x - sec^2 x
cosec^2 x = 1 + cot^2 x
sec^2 x = 1 + tan^2 x
so cosec^2 x - sec^2 x = 1 + cot^2 x - 1 - tan^2 x = cot^2 x - tan^2 x = RHS

Reply 2

Juwel

TheWolf

cot^2x - tan^2x = 4cot2x cosec2x

proof LHS = RHS

can anyone do this?

4cot2xcosec2x = 4(1-tan^2x/2tanx)(1/2sinxcosx)
= (sec^2x/tanx)(1/sinxcosx) = sec^2x/tanxsinxcosx = sec^2x/sin^2x
= 1/sin^2xcos^2x = sin^2x+cos^2x/sin^2xcos^2x
= sin^2x/cos^2x + cos^2x/sin^2x = tan^2x + cot^2x

Hmm...

EDIT: Nope, 1-tan^2x is not sec^2x.

Reply 3

Jonny W

4cot(2x) cosec(2x)
= 4cos(2x) / sin^2(2x)
= [cos^2(x) - sin^2(x)] / [sin^2(x) cos^2(x)]
= cosec^2(x) - sec^2(x)
= [1 + cot^2(x)] - [1 + tan^2(x)]
= cot^2(x) - tan^2(x).

Reply 4

TheWolf

Jonny W

4cot(2x) cosec(2x)
= 4cos(2x) / sin^2(2x)
= [cos^2(x) - sin^2(x)] / [sin^2(x) cos^2(x)]
= cosec^2(x) - sec^2(x)
= [1 + cot^2(x)] - [1 + tan^2(x)]
= cot^2(x) - tan^2(x).

ahh ok thanks guys:

ps i hate all trig p2 questions it takes me longer to do than any other parts

Reply 5

Juwel

TheWolf

ahh ok thanks guys:

ps i hate all trig p2 questions it takes me longer to do than any other parts

Once you do a few and get good at them you'll enjoy unravelling them.

Reply 6

JaF

TheWolf

...ps i hate all trig p2 questions it takes me longer to do than any other parts

The best advice I ever got was to always get rid of the double angles first. Its a lot harder to work towards double angles.

Reply 7

Bezza

JaF

The best advice I eer got was to alwats get rid of the double angles first. Its a lot harder to work towards double angles.

Yeah you need to decide which side is the most complicated and try and simplify it.

Reply 8

Jonny W

TheWolf

ahh ok thanks guys:

ps i hate all trig p2 questions it takes me longer to do than any other parts

Write both sides in terms of powers of sin(x) and cos(x) only. Then if necessary use sin^2 + cos^2 = 1.

Reply 9

TheWolf

yea i think i have to do trig questions all day tommorow lol thats how shit i am at it, i get one in two of them wrong or cant do it at all

Reply 10

Silly Sally

TheWolf

yea i think i have to do trig questions all day tommorow lol thats how shit i am at it, i get one in two of them wrong or cant do it at all

After *attempting* unsuccessfully to do the question you posted, i have realised i can't do trig either. :frown:

Reply 11

Silly Sally

Jonny W

4cot(2x) cosec(2x)
= 4cos(2x) / sin^2(2x)
= [cos^2(x) - sin^2(x)] / [sin^2(x) cos^2(x)]
= cosec^2(x) - sec^2(x)
= [1 + cot^2(x)] - [1 + tan^2(x)]
= cot^2(x) - tan^2(x).