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WebMar 30, 2024 Β· So we use the integrating factor method ππ¦/ππ₯ = ( (1 + π¦^2))/ (tan^ (β1) y β x) This is not of the form ππ¦/ππ₯+ππ¦=π β΄ We need to find ππ₯/ππ¦ ππ₯/ππ¦ = (tan^ (β1)β‘π¦ β π₯)/ (1 + π¦^2 ) ππ₯/ππ¦ = tan^ (β1)β‘π¦/ (1 + π¦^2 ) β π₯/ β¦ WebA viewer wanted me to try the trig equation involving tangent: tan(1/x)=1/tan(x). This seems to be a fake trig identity but it turns out to be a very interes... binding assay examples
Solve the equation tan-1(1+x/1-x)=Ο/4+tan-1x - Brainly.in
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