Simplify \(If\times 4\) to \(4If\).
\[4IfcotA=3thenthevalueof\times \frac{sinA+cosA}{sinA-cosA}\imath s\times 1\]
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[4IfcotA=\frac{3thenthevalueof(sinA+cosA)\imath s\times 1}{sinA-cosA}\]
Regroup terms.
\[4IfcotA=\frac{3tthhnvaluofseee(sinA+cosA)\imath }{sinA-cosA}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[4IfcotA=\frac{3{t}^{2}{h}^{2}nvaluofs{e}^{3}(sinA+cosA)\imath }{sinA-cosA}\]
Regroup terms.
\[4IfcotA=\frac{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}nvaluofs}{sinA-cosA}\]
Multiply both sides by \(sinA-cosA\).
\[4IfcotA(sinA-cosA)=3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}nvaluofs\]
Divide both sides by \(3\).
\[\frac{4IfcotA(sinA-cosA)}{3}={e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}nvaluofs\]
Divide both sides by \({e}^{3}\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3}}{{e}^{3}}=(sinA+cosA)\imath {t}^{2}{h}^{2}nvaluofs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3}}{{e}^{3}}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}}=(sinA+cosA)\imath {t}^{2}{h}^{2}nvaluofs\]
Divide both sides by \(sinA+cosA\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}}}{sinA+cosA}=\imath {t}^{2}{h}^{2}nvaluofs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}}}{sinA+cosA}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)}=\imath {t}^{2}{h}^{2}nvaluofs\]
Divide both sides by \(\imath \).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)}}{\imath }={t}^{2}{h}^{2}nvaluofs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)}}{\imath }\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath }\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath }={t}^{2}{h}^{2}nvaluofs\]
Divide both sides by \({t}^{2}\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath }}{{t}^{2}}={h}^{2}nvaluofs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath }}{{t}^{2}}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}}={h}^{2}nvaluofs\]
Divide both sides by \({h}^{2}\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}}}{{h}^{2}}=nvaluofs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}}}{{h}^{2}}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}}=nvaluofs\]
Divide both sides by \(v\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}}}{v}=naluofs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}}}{v}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}v}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}v}=naluofs\]
Divide both sides by \(a\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}v}}{a}=nluofs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}v}}{a}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}va}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}va}=nluofs\]
Divide both sides by \(l\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}va}}{l}=nuofs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}va}}{l}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}val}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}val}=nuofs\]
Divide both sides by \(u\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}val}}{u}=nofs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}val}}{u}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valu}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valu}=nofs\]
Divide both sides by \(o\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valu}}{o}=nfs\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valu}}{o}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuo}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuo}=nfs\]
Divide both sides by \(f\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuo}}{f}=ns\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuo}}{f}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuof}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuof}=ns\]
Divide both sides by \(s\).
\[\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuof}}{s}=n\]
Simplify \(\frac{\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuof}}{s}\) to \(\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuofs}\).
\[\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuofs}=n\]
Switch sides.
\[n=\frac{4IfcotA(sinA-cosA)}{3{e}^{3}(sinA+cosA)\imath {t}^{2}{h}^{2}valuofs}\]
n=(4*If*cotA*(sinA-cosA))/(3*e^3*(sinA+cosA)*IM*t^2*h^2*v*a*l*u*o*f*s)
