Question

please help solving this identity

Answer 1

?!

Answer 2

\[\cot ^{2}\frac{ u }{ 2 }= \frac{ \csc u + \cot u}{ \csc u -\cot u }\]

Answer 3

I am not sure I what I did is correct because I don't know what else to do for the left side:
\[\cot \frac{ u }{ 2 }* \cot \frac{ u }{ 2 }\]
\[\frac{ 1 }{ \tan \frac{ u }{ 2 } }* \frac{ 1 }{ \tan \frac{ u }{ 2 } }\]
\[\frac{ 1 }{ \frac{ \sin u }{ 1+\cos u} }* \frac{ 1 }{ \frac{ sinu }{ 1+\cos u } }\]

Answer 4

jenny1990 gimme a few secs

Answer 5

\(\bf \cot ^{2}\left(\frac{ u }{ 2 }\right)= \cfrac{ \csc u + \cot u}{ \csc u -\cot u }\\ \quad \\ \quad \\

\cfrac{ \csc u + \cot u}{ \csc u -\cot u }\implies
\cfrac
{\frac{1}{sin(u)}+\frac{cos(u)}{sin(u)}
}
{\frac{1}{sin(u)}-\frac{cos(u)}{sin(u)}
}
\implies \cfrac
{\frac{1+cos(u)}{sin(u)}
}
{\frac{1-cos(u)}{sin(u)}
}\\ \quad \\
\color{red}{\cfrac{1+cos(u)}{1-cos(u)}}\)

Answer 6

now recall the trig identity of \(\bf \cot ^{2}\left(\frac{ u }{ 2 }\right)= \cfrac{ \csc u + \cot u}{ \csc u -\cot u }\\ \quad \\ \quad \\
tan\left(\frac{u}{2}\right)=\sqrt{\cfrac{1-cos(u)}{1+cos(u)}}\implies tan^2\left(\frac{u}{2}\right)=\cfrac{1-cos(u)}{1+cos(u)}\\ \quad \\
\textit{now let us raise both sides by -1}\\ \quad \\
\left(tan^2\left(\frac{u}{2}\right)\right)^{-1}=\left(\cfrac{1-cos(u)}{1+cos(u)}\right)^{-1}\implies \cfrac{1}{tan^2\left(\frac{u}{2}\right)}=\cfrac{1+cos(u)}{1-cos(u)}\\ \quad \\
\textit{recall that }\quad cot(\theta) = \cfrac{1}{tan(\theta)}\qquad thus\\ \quad \\
\color{red}{cot^2\left(\frac{u}{2}\right)=\cfrac{1+cos(u)}{1-cos(u)}}\)

Answer 7

thank u so much, I was completely lost

Answer 8

yw

Answer 9

hey is it ok if I ask another question? it is about the same thing I just that I am not sure If the way I did it is correct