Is 1-cos(θ)=(sin^2(θ))/(1+cos(θ)) ?

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

prove 1-cos(θ)=(sin^2(θ))/(1+cos(θ))

prove

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

True

Solution steps

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

Manipulating left side

1 - cos (θ)

Multiply by

\frac{1 + cos ( θ )}{1 + cos ( θ )}

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

Apply Difference of Two Squares Formula:

θ^{2} - y^{2} = (θ + y) (θ - y)

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

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - cos^{2} (θ) = sin^{2} (θ)

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{sin ( 2 x )}{sin ( x )} = \frac{2}{sec ( x )}

p ro v e \frac{2}{1 + cos ( 2 x )} = sec^{2} (x)

p ro v e cos (2 x) = - 2 sin (x^{2}) + 1

p ro v e tan (x) (cos (x) + cot (x)) = sin (x) + 1

p ro v e 1 + tan^{2} (6 0^{\circ}) = sec^{2} (6 0^{\circ})

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