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

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

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prove cot(x)=(sin(2x))/(1-cos(2x))

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

1 - 1 = 0

= 2 sin^{2} (x)

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

sin (x)

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (- \frac{5 π}{6})

tan (\frac{5 π}{8})

sin (10 0^{\circ})

sec (6 0^{\circ})

tan (27 0^{\circ})

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