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Splice1108
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Find y[itex]^{I}[/itex] (sinxsecx)/1+xtanx The supplied answer is 1/(1+xtanx)[itex]^{2}[/itex]
I got stuck with an extra x on top at the end. Where did I mess up at?
y[itex]^{I}[/itex](sinxsecx)/1+xtanx = [1+xtanx*f[itex]^{I}[/itex](sinxsecx)-sinxsecx*f[itex]^{I}[/itex](1+xtanx)]/(1+xtanx)[itex]^{2}[/itex] = [1+xtanx((cos/1)*(1/cos))-((cos/1)*(1/cos))*(0-1*(-cos/sin))]/(1xtanx)[itex]^{2}[/itex]=[1+x*(sinx/cosx)*(cosx/sinx)*(-1)+1]/(1+xtanx)[itex]^{2}[/itex]=(1+(x)(1)(-1)+1)/(1+xtanx)[itex]^{2}[/itex]
I realize how much of a mess it is. I suspect it's something very simple that I missed.
I got stuck with an extra x on top at the end. Where did I mess up at?
y[itex]^{I}[/itex](sinxsecx)/1+xtanx = [1+xtanx*f[itex]^{I}[/itex](sinxsecx)-sinxsecx*f[itex]^{I}[/itex](1+xtanx)]/(1+xtanx)[itex]^{2}[/itex] = [1+xtanx((cos/1)*(1/cos))-((cos/1)*(1/cos))*(0-1*(-cos/sin))]/(1xtanx)[itex]^{2}[/itex]=[1+x*(sinx/cosx)*(cosx/sinx)*(-1)+1]/(1+xtanx)[itex]^{2}[/itex]=(1+(x)(1)(-1)+1)/(1+xtanx)[itex]^{2}[/itex]
I realize how much of a mess it is. I suspect it's something very simple that I missed.
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