Is cot^2(v/2)=((sec(v)+1))/((sec(v)-1)) ?

The answer to whether cot^2(v/2)=((sec(v)+1))/((sec(v)-1)) is True

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prove cot^2(v/2)=((sec(v)+1))/((sec(v)-1))

prove

cot^{2} (\frac{v}{2}) = \frac{( sec ( v ) + 1 )}{( sec ( v ) - 1 )}

True

Solution steps

cot^{2} (\frac{v}{2}) = \frac{sec ( v ) + 1}{sec ( v ) - 1}

Let:

u = \frac{v}{2}

cot^{2} (u) = \frac{sec ( 2 u ) + 1}{sec ( 2 u ) - 1}

Prove

cot^{2} (u) = \frac{sec ( 2 u ) + 1}{sec ( 2 u ) - 1} :

True

Therefore

cot^{2} (\frac{v}{2}) = \frac{sec ( v ) + 1}{sec ( v ) - 1}

We showed that the two sides could take the same form

\Rightarrow True

csc (θ) (1 - cos (θ)) (1 + cos (θ)) = sin (θ)

sec (θ) - \frac{1}{sec ( θ )} = sin (θ) tan (θ)

csc^{2} (θ) + sec^{2} (θ) = csc^{2} (θ) sec^{2} (θ)

sin^{2} (x) = tan (x) cot (x) - cos^{2} (x)

sin^{2} (x) - cos^{2} (x) = 1

Is cot^2(v/2)=((sec(v)+1))/((sec(v)-1)) ?
The answer to whether cot^2(v/2)=((sec(v)+1))/((sec(v)-1)) is True