How to verify this trig identiy: (sin(x)-cos(x))^2/(sin(x) cos(x))= sec(x) csc(x) -2?

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answers:
Como say: Assuming that should read as :-
( sin x - cosx )² / ( sin x cos x )
( sin²x - 2 sin x cos x + cos²x ) / (sin x cos x )
( 1 - 2 sin x cos x ) / sin x
csc x - 2 cos x
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la console say: = [sin(x) - cos(x)]² / [sin(x).cos(x)] → recall: (a - b)² = a² - 2ab + b²

= [sin²(x) - 2.sin(x).cos(x) + cos²(x)] / [sin(x).cos(x)] → recall: sin²(x) + cos²(x) = 1

= [1 - 2.sin(x).cos(x)] / [sin(x).cos(x)] → recall: (a - b)/c = (a/c) - (b/c)

= [1/sin(x).cos(x)] - [2.sin(x).cos(x)/sin(x).cos(x)]

= [1/sin(x).cos(x)] - 2

= [1/sin(x)].[1/cos(x)] - 2 → you know that: sec(x) = 1/cos(x)

= [1/sin(x)].sec(x) - 2 → you know that: csc(x) = 1/sin(x)

= csc(x).sec(x) - 2
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MyRank say: sinx - cos²x / sinxcosx = secx cosecx - 2

Sinx = (1-sin²x) / sinx cosx

Sinx = 1 + sin²x / sinxcosx = sin²x + sinx - 1/sinx cosx

= tanx + secx - cosecx.secx.
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Lisa A say: Start substituting and simplifying. If you get the same thing on both sides of the equation, or if you get an identity that you have already proven, or if you get the Pythagorean theorem, then you have proved the identity.



Why does sin(x)^2 + cos(x)^2 = 1 ?
Because you can quickly and easily manipulate it into the Pythagorean theorem,
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Captain Matticus, LandPiratesInc say: (a - b)^2 = a^2 - 2ab + b^2
sin(x)^2 + cos(x)^2 = 1
1/sin(x) = csc(x)
1/cos(x) = sec(x)
sin(2x) = 2 * sin(x) * cos(x)

(sin(x) - cos(x))^2 / (sin(x) * cos(x)) =>
(sin(x)^2 - 2 * sin(x) * cos(x) + cos(x)^2) / (sin(x) * cos(x)) =>
(1 - 2 * sin(x) * cos(x)) / (sin(x) * cos(x)) =>
(1 / (sin(x) * cos(x))) - 2 * sin(x) * cos(x) / (sin(x) * cos(x)) =>
(1/sin(x)) * (1/cos(x)) - 2 =>
csc(x) * sec(x) - 2
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new_bumble_bee say: Some silly old hag is what they used to tag it as but, on a more serious note : can't remember much mathematics - got a brain injury so I can sometimes even have trouble remembering most words
And even if I could help you - I would recommend you do your own work anyway

---- also noticed the other answers as well
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