Is 1-tan^2(θ/2)=(2cos(θ))/(1+cos(θ)) ?

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

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

prove

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

True

Solution steps

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

Let:

u = \frac{θ}{2}

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

Prove

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

True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( x ) - csc ( x )}{csc ( x )} = - cos^{2} (x)

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

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

cot^{2} (θ) csc^{2} (θ) - cot^{2} (θ) = cot^{4} (θ)

sin (x - π) = - sin (x)

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