Is cot(θ)=(sin(2θ))/(1-cos(2θ)) ?

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

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prove cot(θ)=(sin(2θ))/(1-cos(2θ))

prove

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

True

Solution steps

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

Manipulating right side

\frac{sin ( 2 θ )}{1 - cos ( 2 θ )}

Rewrite using trig identities

\frac{sin ( 2 θ )}{1 - cos ( 2 θ )}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

1 - (1 - 2 sin^{2} (θ)) : 2 sin^{2} (θ)

1 - (1 - 2 sin^{2} (θ))

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

- (1 - 2 sin^{2} (θ))

Distribute parentheses

= - (1) - (- 2 sin^{2} (θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (θ)

= 1 - 1 + 2 sin^{2} (θ)

1 - 1 = 0

= 2 sin^{2} (θ)

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

sin (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

\frac{cos ( x )}{sin ( x )} = cot (x)

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

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