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

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

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

prove

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

True

Solution steps

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

Manipulating left 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 )}

We showed that the two sides could take the same form

\Rightarrow True

\frac{( cos ( a - b ) )}{sin ( a ) sin ( b )} = cot (a) cot (b) + 1

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

sin^{2} (x) = 3 cos^{2} (x)

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

2 cos (x) (sin (3 x) - sin (x)) = sin (4 x)

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