Is sin^2(x/2)=(sec(x)-1)/(2sec(x)) ?

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

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prove sin^2(x/2)=(sec(x)-1)/(2sec(x))

prove

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

True

Solution steps

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

Let:

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

We showed that the two sides could take the same form

\Rightarrow True

cos (θ) + tan (θ) sin (θ) = sec (θ)

cot^{2} (x) + 5 = csc^{2} (x) + 4

\frac{sin ( θ )}{sin ( θ ) - 1} = \frac{1}{1 - csc ( θ )}

cot (x) (sec (x) - cos (x)) = sin (x)

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

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