Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 years ago

Trig Identities Help Please!?

sin(x) - sin(x)

-------- -------- =tan^2(x)

csc(x)-1 cot^2(x)

Thanks in advance!

1 Answer

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  • NCS
    Lv 7
    3 years ago
    Favorite Answer

    Let me re-write it as

    sin(x)/(csc(x)-1) - sin(x)/cot²(x) = tan²x

    Dealing with the first term on the LHS:

    csc(x) = 1/sin(x), so

    LHS1 = sin(x)/(csc(x) - 1) * sin(x)/sin(x) = sin²(x) / (1 - sin(x))

    ... multiply by (1+sin(x)) / (1+sin(x))

    LHS1 = sin²(x) / (1 - sin(x)) * (1 + sin(x)) / (1 + sin(x))

    LHS1 = sin²(x) * (1 + sin(x)) / (1 - sin²(x))

    ... and 1 - sin²(x) = cos²(x), so

    LHS1 = sin²(x) * (1 + sin(x)) / cos²(x) = tan²(x) * (1 + sin(x))

    Dealing with the second term on the LHS:

    1/cot²(x) = tan²(x), so

    LHS2 = tan²(x) * sin(x)

    So

    LHS = LHS1 - LHS2 = tan²(x)*(1 + sin(x)) - tan²(x)*sin(x)

    LHS = tan²(x) * (1 + sin(x) - sin(x)) = tan²(x) = RHS

    Hope this helps!

    Source(s): Please revisit https://answers.yahoo.com/question/index?qid=20161... Thanks in advance
    • Captivity3 years agoReport

      Sorry, I didn't see you commented, its csc(x) - 1

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