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Trigonometric Identities

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I need a reply asap!
How can this identity be proved?
(sin^2 x - cos^2 x) / (1 + 2 sinx cosx) == (tan x - 1)/ (tan x + 1) ????
 

XPFMember

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Actually i haven't done these type of questions yet...but i just went through sum notes on the net...and i have found out a few things...i dont know whether u know or not cuz i am unaware of what's there in Add Maths!!
so sin^2 x = cos^2 x - cos (2x)
so we can write sin^2 x - cos^2 x => [cos^2 x - cos (2x)] - cos^2 x => -cos (2x)


and also 2 sinx cosx => sin 2x or 2 sinx cosx=> (2 tan x ) /(1 + tan^2 x)


I dont know if this information is helpful or not...bcoz since i havent done these type of questions i cudnt do it...may be in the next few days will be able to tell as we started trignometry in school today only...(AS Maths)
 
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RHS=tan x - 1/ tan x + 1
= (sin x/cos x ) - 1 /(sin x/cos x) + 1 {since tan x = sin x/cos x)
= (sin x - cos x)/cos x / (sin x + cos x)/ cos x { cos x as a common denominator}
= (sin x - cos x) / cos x * cos x /(sin x + cos x) { cos x is cancelled out}
= (sin x - cos x) / (sin x + cos x) { multiply by (sin x + cos x) in numerator and denominator}
= [sin x(sin x + cos x) - cos x(sin x + cos x)] / (sin x + cos x) ^ 2
= (sin ^ 2 x - cos ^2 x) / sin^2 x + 2sin x cos x + cos ^2 x
= sin ^2 x - cos ^2 x / 1 + 2 sin x cos x { since sin ^2 x + cos ^2x = 1}
= LHS
hence proved
u can do it the other way round as well... starting from the LHS
 
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Thank you !the question was quite simple! sometimes one just doesn't have the right mind for Add-maths!!!
 
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Lol... Sadly, examiners don't consider your liking before making questions :p
 
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