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TheWolf

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

proof LHS = RHS

can anyone do this?

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

TheWolf

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

proof LHS = RHS

can anyone do this?

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.

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.

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).

= 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).

Does sin^2(2x) = sin^2(x)cos^2(x) ?

Which rule is that?

ZJuwelH

sin^2(2x) = (sin2x)^2 = (2sinxcosx)^2 = 4sin^2xcos^2x, the 4 is cancelled by the 4 outside.

AHHHH!!! Thanks!!! All unconfuddled now!!!

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can someone please explain what principle domain is and why the answer is a not c?Maths

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