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

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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 right side

\frac{sin ^{2} ( θ )}{1 - cos ( θ )}

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{1 - cos ( θ )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

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

Simplify

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

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

Factor

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

1 - cos^{2} (θ)

Factor out common term

- 1

= - (cos^{2} (θ) - 1)

Factor

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

cos^{2} (θ) - 1

Rewrite

1

1^{2}

= cos^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (θ) + 1) (cos (θ) - 1)

= - (cos (θ) + 1) (cos (θ) - 1)

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

Cancel

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

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

- cos (θ) + 1 = - (cos (θ) - 1)

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

Refine

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

Cancel the common factor:

cos (θ) - 1

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= 1 + cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

cot (x) - sin (2 x) = cot (x) cos (2 x)

sin (2 θ) = 2 sin (2 θ) cos (2 θ)

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

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

\frac{1}{cos ( θ )} - cos (θ) = sin (θ) tan (θ)

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