Is sin^2(θ/2)=(sec(θ)-1)/(2sec(θ)) ?

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

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prove sin^2(θ/2)=(sec(θ)-1)/(2sec(θ))

prove

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

True

Solution steps

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

Let:

u = \frac{θ}{2}

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

Prove

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

True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

cot (t) csc (t) = \frac{csc ^{2} ( t ) - 1}{cos ( t )}

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

\frac{1 - cos ( x )}{1 + cos ( x )} = tan^{2} (\frac{x}{2})

sin (\frac{π}{3}) = \frac{3}{2}

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

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