# Prove this trig identtiy:?

sec^2-sec^2 = tan^2x - tan^2y

Update:

OOPS I MEANT:

sec^2x - sec^2y = tan^2x - tan^2y

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- gudspelingLv 71 decade agoFavorite Answer
LHS

= sec²(x) - sec²(y)

= (1+tan²(x)) - (1+tan²(y))

= 1 + tan²(x) - 1 - tan²(y)

= tan²(x) - tan²(y)

= RHS

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- Anonymous1 decade ago
use sec^2 x = 1 + tan^2 x

sec^2x - sec^2y = 1 + tan^2 x - (1 + tan^2 y) = tan^2 x - tan^2 y

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- ireadlotsLv 61 decade ago
If you don't remember the sec/tan identity (I could never remember them all), you can get there logically:

Tan = sin/cos, so

RHS = sin^2(x)/Cos^2(x) - Sin^2(y)/Cos^2(y)

= (1-Cos^2(x))/Cos^2(x) - (1-Cos^2(y))/Cos^2(y)

Put them over a common denominator, and you get a top line of:

[Cos^2(y)-Cos^2(x)Cos^2(y)

-Cos^2(x)+Cos^2(x)Cos^2(y)

which gives you

=..[cos^2(y)-Cos^2(x)]

_________________

[cos^2(x)Cos^2(y)]

Cancelling out, that gives you

= 1/Cos^2(x) - 1/Cos^2(y)

= sec^2(x) - sec^2(y)

= LHS

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