Is (sin^2(θ))/(1-cos(θ))-1=cos(θ) ?

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

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prove (sin^2(θ))/(1-cos(θ))-1=cos(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

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 )} - 1

Simplify

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

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

Cancel

\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 - 1

1 - 1 = 0

= cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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