Is tan^2(θ/2)=(sec(θ)-1)/(sec(θ)+1) ?

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

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prove tan^2(θ/2)=(sec(θ)-1)/(sec(θ)+1)

prove

tan^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{sec ( θ ) + 1}

True

Solution steps

tan^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{sec ( θ ) + 1}

Let:

u = \frac{θ}{2}

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

Prove

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

True

Therefore

tan^{2} (\frac{θ}{2}) = \frac{sec ( θ ) - 1}{sec ( θ ) + 1}

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( x )}{tan ( x )} = csc (x)

\frac{1 - sin ( v )}{cos ( v )} + \frac{cos ( v )}{1 - sin ( v )} = 2 sec (v)

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

sec (x) = \frac{1 + csc ( x )}{cot ( x ) + cos ( x )}

9 sec^{2} (θ) - 5 tan^{2} (θ) = 5 + 4 sec^{2} (θ)

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