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AaronAndyson

  • one year ago

sec^2(A) + cosec^2(A) = sec^2(A).cosec^2(A) @michele_liano

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  1. AaronAndyson
    • one year ago
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    @Michele_Laino

  2. Michele_Laino
    • one year ago
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    we have to apply these identities: \[\sec A = \frac{1}{{\cos A}},\quad \csc A = \frac{1}{{\sin A}}\]

  3. AaronAndyson
    • one year ago
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    ?

  4. imqwerty
    • one year ago
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    have u tried to solve the expression?? try converting the terms into sin nd cos form nd then take the LCM nd solve

  5. Michele_Laino
    • one year ago
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    left side becomes: \[\sec A + {\left( {\csc A} \right)^2} = \frac{1}{{\cos A}} + {\left( {\frac{1}{{\sin A}}} \right)^2} = ...\] please continue

  6. AaronAndyson
    • one year ago
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    i'm confused

  7. Michele_Laino
    • one year ago
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    I think that there is a typo into your original expression, please check

  8. AaronAndyson
    • one year ago
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    fixed

  9. Michele_Laino
    • one year ago
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    ok! so left side becomes: \[{\left( {\sec A} \right)^2} + {\left( {\csc A} \right)^2} = {\left( {\frac{1}{{\cos A}}} \right)^2} + {\left( {\frac{1}{{\sin A}}} \right)^2} = ...\]

  10. imqwerty
    • one year ago
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    ok m calling sinA = x nd cosA=y 1/y^2 +1/x^2 take LCM we get - (x^2 + y^2)/(xy)^2 we know that sin^2A+cos^2A = 1 therefore the numerator = 1 nd we r left with 1/(xy)^2 =sec^A cosec^A

  11. AaronAndyson
    • one year ago
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    sin^2(A)+cos^2(A)/cos(A)sin(A) = RHS

  12. imqwerty
    • one year ago
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    yes

  13. AaronAndyson
    • one year ago
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    1/sin(A)cos(A) = sec(A)cosec(A) = RHS

  14. Michele_Laino
    • one year ago
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    more precisely, it is: sin^2(A)+cos^2(A)/cos(A)^2 sin(A)^2 = RHS

  15. AaronAndyson
    • one year ago
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    ok!thanks...

  16. AaronAndyson
    • one year ago
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    i nedd more help *need

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