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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - cos^{2} (t)

Factor out common term

- 1

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

Factor

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

cos^{2} (t) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (cos (t) + 1) (cos (t) - 1)

= - (cos (t) + 1) (cos (t) - 1)

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

Cancel

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

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

- cos (t) + 1 = - (cos (t) - 1)

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

Refine

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

Cancel the common factor:

cos (t) - 1

= cos (t) + 1

= cos (t) + 1

= cos (t) + 1

= cos (t) + 1

= 1 + cos (t)

We showed that the two sides could take the same form

\Rightarrow True

cos (3 x) = cos (x) (1 - 4 sin^{2} (x))

csc (x) cos (x) = cos^{2} (x)

tan (A) + cot (A) = \frac{2}{sin ( 2 A )}

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

cos (θ) \cdot tan (θ) \cdot csc (θ) = 1

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