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

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

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prove (sin^2(x))/(1-cos(x))=1+cos(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - cos^{2} (x)

Factor out common term

- 1

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

Factor

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

cos^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (cos (x) + 1) (cos (x) - 1)

= - (cos (x) + 1) (cos (x) - 1)

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

Cancel

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

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

- cos (x) + 1 = - (cos (x) - 1)

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

Refine

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

Cancel the common factor:

cos (x) - 1

= cos (x) + 1

= cos (x) + 1

= cos (x) + 1

= cos (x) + 1

= 1 + cos (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (x + y) + sin (x - y) = 2 sin (x) cos (y)

cos (2 θ) = cos^{2} (θ) - sin^{2} (θ)

sec (θ) - cos (θ) = tan (θ) sin (θ)

\frac{csc ( θ ) + sec ( θ )}{tan ( θ ) + 1} = csc (θ)

cot (x) cos (x) + sin (x) = csc (x)

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