Prove that tanx + cotx = (secx)(cscx).?

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  • Anonymous
    1 decade ago
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    left part

    tan x = sinx / cosx

    cot x = cosx / sinx

    sin²x + cos²x = 1

    ((sinx)² + (cosx)²) / (cosx*sinx)

    = 1 / (cosx* sinx)

    right part

    sec = 1 / cosx

    csec = 1 / sinx

    1 / (sinx*cosx)

    so left = right

  • 1 decade ago

    Write everything in terms of sin and cos:

    (sinx/cosx)+(cosx/sinx)=(1/cosx)(1/sinx)

    Find a common denominator of cosx*sinx on the left side and simplify the right side:

    (sin^2x+cos^2x)/(sinx*cosx)=1/(sinx*cosx)

    Realize that the numerator of the left side is an identity that equals 1. Thus:

    1/(sinx*cosx)=1/(sinx*cosx)

  • 3 years ago

    replace tanx to sinx/cosx then replace cotx to cosx/sinx replace secx to a million/cosx and cscx to a million/sinx sinx/cosx + cosx/sinx=a million/cosx(a million/sinx) while u upload u upload or subtract u could desire to get consumer-friendly denominators for that edge of the equation so situations sinx/cosx by potential of sinx and situations cosx/sinx by potential of cosx sin^2x +cos^2x/cosxsinx=a million/cosx(a million/sinx) sin^2x + cos^2x = a million so replace it for a million a million/cosxsinx=a million/cosx(a million/sinx) so now for the different edge of the equation diverse for the duration of with a million/cosx(a million/sinx) and u get..... a million/cosxsinx=a million/cosxsinx =]

  • 1 decade ago

    tan x + cot x = sin x/cos x + cos x/sin x =

    (sin^2 x/ (cos x *sin x)) + (cos^2 x/(sin x* cos x) =

    (sin^2 x + cos^2 x) / (cos x * sin x) =

    1/ (cos x * sin x) =

    1/ cos x * 1/sin x =

    sec x * csc x

    Source(s): Geometry class
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