help 12

If you replace sin x by sqrt(1- cos²x) then further manipulation gives cos(A) = 2e^x/(e^2x + 1)

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If tanA+secA=e^x , then cosA=?

\(\frac{sinA}{cosA}+\frac{1}{cosA}=e^x\\ \frac{(sinA)+1}{cosA}=e^x\\ \frac{(sinA)+1}{cosA}=e^x\\ (sin(A)+1)e^{-x}=cosA\)

\(Cos(A)=\frac{1+sin(A)}{e^x}\)

I am not sure if you can do more than this.  Nothing is jumping out at me ://

Lets see

\(CosA=\frac{1+sinA}{e^x}\\ CosA=\frac{1+(1-cos^2A)^{0.5}}{e^x}\\ e^xCosA=1+(1-cos^2A)^{0.5}\\ e^xCosA-1=(1-cos^2A)^{0.5}\\ e^{2x}Cos^2A-2e^xCosA+1=1-Cos^2A\\ e^{2x}Cos^2A+Cos^2A-2e^xCosA=0\\ (e^{2x}+1)Cos^2A-2e^xCosA=0\\ (e^{2x}+1)Cos^2A-2e^xCosA=0\\ cosA[(e^{2x}+1)CosA-2e^x]=0\\ cosA=0 \qquad or \qquad (e^{2x}+1)CosA-2e^x=0\\ cosA=0 \qquad or \qquad CosA=\frac{2e^x}{ e^{2x}+1 }\\\)

tanA + secA = e^x

tanA  = e^x - secA     square both sides

tan^2A  = e^[2x] - 2secA*[e^x] + sec^2A

tan^2A  = e^[2x] - 2secA*[e^x] + tan^2A + 1

0 = e^[2x] - 2secAe^x + 1

2secAe^x = e^[2x] + 1

secA  = (e^[2x] + 1 )/ (2e^[2x])

cosA  = (2e^[2x] ) / ( e^[2x] + 1 )

Note that Melody gets an extraneous solution of cosA = 0 ......but this cannot be so, because this would imply that secA  = 1/cosA  = 1/0     and this would be undefined in the original assumption

Thanks Chris      

I should have seen that