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2[sin6A+cos6A]-3[sin4A+cos4A]+1=0

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2[sin6A+cos6A]-3[sin4A+cos4A]+1=0

    • Posted by Kajal Dogra 6 years, 10 months ago

    CBSE > Class 10 > Mathematics
    • 3 answers

    Ujjwal Gautam 6 years, 10 months ago

    Sorry, put these in LHS

    0Thank You

    Ujjwal Gautam 6 years, 10 months ago

    Sin^6A+cos^6A=(sin^2A+cos^2A)^3-3sin^2Acos^2A(sin^2A+cos^2A)=1-3sin^2Acos^2 And. Sin^4A+cos^4A=(sin^2A+cos^2A)^2-2sin^2Acos^A=1-2sin^2Acos^2A Put these values in Rhs

    0Thank You

    Lavish Pareek 6 years, 10 months ago

    2[sinA +sin (90-A)] - 3[ sin4A + sin(90-4A)] +1 2[1] -3[1] +1 =0

    0Thank You

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