Prove this trig identtiy:?

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

Update:

OOPS I MEANT:

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

3 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    LHS

    = sec²(x) - sec²(y)

    = (1+tan²(x)) - (1+tan²(y))

    = 1 + tan²(x) - 1 - tan²(y)

    = tan²(x) - tan²(y)

    = RHS

    • Login to reply the answers
  • Anonymous
    1 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

    • Login to reply the answers
  • 1 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

    • Login to reply the answers
Still have questions? Get your answers by asking now.